# The first prime number after the 8th Fermat number

In this post, we discuss a primality testing exercise involving the eighth Fermat number. A Fermat number is of the form $F_n=2^{2^n}+1$ where $n$ is any nonnegative integer. We search for the first prime number that is greater than $F_8$. The basic idea is to search for the first probable prime base 2 among the odd numbers after $F_8$. Once the first probable prime base 2 is identified, we apply the Miller-Rabin primality test to confirm that it is a prime number. At the outset of this exercise, we did not know how many numbers we had to check before reaching the first prime number.

The first five Fermat numbers $F_0$, $F_1$, $F_2$, $F_3$ and $F_4$ are the only Fermat numbers that are known to be prime (it was conjectured by Fermat that all Fermat numbers are prime). It is unknown whether there exists prime Fermat number beyond $F_4$. What is clear, however, is that all the higher Fermat numbers that were studied turn out to be composite. The 8th Fermat number $F_8$ has 78 decimal digits with two factors with 16 and 62 digits (it was factored in 1961). The largest Fermat number that has been completely factored (as of the writing of this post) is $F_{11}$ which has 617 decimal digits. Many Fermat numbers larger than $F_{11}$ have been partially factored.

___________________________________________________________________

The basic approach

The following is the number $2^{256}$, which has 78 decimal digits.

$2^{256}=$
11579208923731619542357098500868790785326998466564
0564039457584007913129639936

Define $P_j=2^{256}+j$ where $j$ is an odd positive integer, i.e., $j=1,3,5,7,\cdots$. The exercise is to find the smallest $j$ such that $P_j$ is a prime number. According to Euclid’s proof that there are infinitely many prime numbers, such a $P_j$ is sure to exist. Just that we do not know at the outset how far we have to go to find it. Of course, $P_1$ is the 8th Fermat number, which is a composite number with two prime factors with 16 and 62 decimal digits. So the search starts with $j=3$.

The key is to do the following two quick checks to eliminate composite numbers so that we can reach a probable prime as quickly as possible.

• For any given $P_j$, the first step is to look for small prime factors, i.e., to factor $P_j$ using prime numbers less than a bound $B$. If a small prime factor is found, then we increase $j$ by 2 and start over. Note that we skip any $P_j$ where the sum of digits is divisible by 3. We also skip any $P_j$ that ends with the digit 5.
• If no small factors are found, then compute the congruence $2^{P_j-1} \ (\text{mod} \ P_j)$. If the answer is not congruent to 1, then we know $P_j$ is composite and work on the next number. If $2^{P_j-1} \equiv 1 \ (\text{mod} \ P_j)$, then $P_j$ is said to be a probable prime base 2. Once we know that a particular $P_j$ is a probable prime base 2, it is likely a prime number. To further confirm, we apply the Miller-Rabin primality test on that $P_j$.

In the first check, we check for prime factors among the first 100 odd prime numbers (i.e. all odd primes up to and including 547).

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Searching the first probable prime

At the outset, we did not know how many numbers we will have to check. Since there can be a long gap between two successive prime numbers, the worse fear is that the number range we are dealing with is situated in such a long gap, in which case we may have to check thousands of numbers (or even tens of thousands). Luckily the search settles on a probable prime rather quickly. The magic number is 297. In other words, for the number

$P_{297}=2^{256}+297$

, we find that $2^{P_{297}-1} \equiv 1 \ (\text{mod} \ P_{297})$. Thus $P_{297}$ is a probable prime in base 2. The following shows the decimal digits of $P_{297}$.

$P_{297}=$
11579208923731619542357098500868790785326998466564
0564039457584007913129640233

To further give a sense of how the magic number $P_{297}$ is reached, the following table lists the 25 calculations leading to the magic number.

$\left[\begin{array}{rrrrrrr} j & \text{ } & \text{last 5 digits of } P_j & \text{ } & \text{least factor of } P_j & \text{ } & 2^{P_j-1} \ \text{mod} \ P_j \\ \text{ } & \text{ } & \text{ } \\ 259 & \text{ } & 40195 & \text{ } & 5 & \text{ } & \text{ } \\ 261 & \text{ } & 40197 & \text{ } & * & \text{ } & \not \equiv 1 \\ 263 & \text{ } & 40199 & \text{ } & 3 & \text{ } & \text{ } \\ 265 & \text{ } & 40201 & \text{ } & * & \text{ } & \not \equiv 1 \\ 267 & \text{ } & 40203 & \text{ } & * & \text{ } & \not \equiv 1 \\ 269 & \text{ } & 40205 & \text{ } & 3 & \text{ } & \text{ } \\ 271 & \text{ } & 40207 & \text{ } & 7 & \text{ } & \text{ } \\ 273 & \text{ } & 40209 & \text{ } & * & \text{ } & \not \equiv 1 \\ 275 & \text{ } & 40211 & \text{ } & 3 & \text{ } & \text{ } \\ 277 & \text{ } & 40213 & \text{ } & 11 & \text{ } & \text{ } \\ 279 & \text{ } & 40215 & \text{ } & 5 & \text{ } & \text{ } \\ 281 & \text{ } & 40217 & \text{ } & 3 & \text{ } & \text{ } \\ 283 & \text{ } & 40219 & \text{ } & 13 & \text{ } & \text{ } \\ 285 & \text{ } & 40221 & \text{ } & 7 & \text{ } & \text{ } \\ 287 & \text{ } & 40223 & \text{ } & 3 & \text{ } & \text{ } \\ 289 & \text{ } & 40225 & \text{ } & 5 & \text{ } & \text{ } \\ 291 & \text{ } & 40227 & \text{ } & 23 & \text{ } & \text{ } \\ 293 & \text{ } & 40229 & \text{ } & 3 & \text{ } & \text{ } \\ 295 & \text{ } & 40231 & \text{ } & 71 & \text{ } & \text{ } \\ 297 & \text{ } & 40233 & \text{ } & * & \text{ } & \equiv 1 \end{array}\right]$

The first number is in the table $P_{259}$ ends in a 5 and is thus composite. The third number $P_{263}$ is composite since the sum of the digits of its is divisible by 3. The third column of the above table shows the least prime factor below 547 (if one is found). An asterisk in the third column means that none of the prime numbers below 547 is a factor. For such numbers, we compute the modular exponentiation $2^{P_j-1} \ (\text{mod} \ P_j)$.

In the above table, 4 of the asterisks lead to the result $2^{P_j-1} \not \equiv 1 \ (\text{mod} \ P_j)$. These numbers $P_j$ are thus composite. For example, for $P_{273}$, the following is the result:

$2^{P_{273}-1} \ (\text{mod} \ P_{273}) \equiv$
55365573520609500639906523255562025480037454102798
631593548187358338340281435

The last number $P_{297}$ in the table is a probable prime base 2 since our calculation shows that $2^{P_{297}-1} \equiv 1 \ (\text{mod} \ P_{297})$. Being a probable prime to base 2 is actually very strong evidence that the number is a prime number. We want even stronger evidence that $P_{297}$ is a prime. For example, we can carry out the Miller-Rabin test in such a way that the probability of mistaking a composite number as prime is at most one in a septillion! A septillion is the square of a trillion. A trillion is $10^{12}$. Thus a septillion is $10^{24}$. One in a septillion is for all practical purposes zero. But if one wants more reassurance, one can always run the Miller-Rabin test with more bases.

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The Miller-Rabin primality test

The Miller-Rabin test is a variant of the Fermat test because Miller-Rabin still relies on Fermat’s little theorem. But Miller-Rabin uses Fermat’s little theorem in such a way that it eliminates the issue of the Fermat test mistakenly identifying Carmichael numbers as prime.

Given an odd positive integer whose “prime or composite” status is not known, the Miller-Rabin test will output “composite” or “probable prime”. Like the Fermat test, the Miller-Rabin test calculates $a^{n-1} \ (\text{mod} \ n)$ for several values of $a$. But the test organizes the congruence $a^{n-1} \ (\text{mod} \ n)$ a little differently to capture additional information about prime numbers.

Here’s how to set up the calculation for Miller-Rabin. Because $n$ is odd, $n-1$ is even. We can factor $n-1$ as a product of a power of 2 and an odd number. So we have $n-1=2^k \cdot q$ where $k \ge 1$ and $q$ is odd ($q$ may not be prime). Then we calculate the following sequence:

$a^q, \ a^{2 \cdot q}, \ a^{2^2 \cdot q}, \cdots, a^{2^{k-1} \cdot q}, \ a^{2^{k} \cdot q} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

The first term in (1) can be calculated using the fast powering algorithm (using the binary expansion of $q$ to convert the calculation of $a^q$ into a series of squarings and multiplications). Each subsequent term is then the square of the preceding term. The last term is of course $a^{n-1}$. Each squaring or multiplication is reduced modulo $n$. The Miller-Rabin test is based on the following property of prime numbers:

Theorem 1
Let $n$ be an odd prime number such that $n-1=2^k \cdot q$ where $k \ge 1$ and $q$ is odd. Let $a$ be a positive integer not divisible by $n$. Then the following two conditions are true about the sequence (1).

• At least one term in the sequence (1) is congruent to 1 modulo $n$.
• Either the first term in (1) is congruent to 1 modulo $n$ or the term preceding the first 1 is congruent to -1 modulo $n$.

How the Miller-Rabin test works
Suppose that the “prime or composite” status of an odd integer $n$ is not known. If both conditions in the above theorem are satisfied with respect to the number $a$, then $n$ is said to be a strong probable prime in base $a$. If a strong probable prime in base $a$ happens to be composite, then it is said to be a strong pseudoprime in base $a$. In other words, a strong pseudoprime is a composite number that possesses a prime-like property, namely it satisfies the two conditions in Theorem 1 with respect to one base $a$.

The test procedure of Miller-Rabin is to check whether $n$ is a strong probable prime to several bases that are randomly chosen. The following determines the outcome of the test:

• If $n$ is not a strong probable prime in one of the chosen bases, then $n$ is proved to be composite.
• If $n$ is shown to be a strong probable prime in all the chosen bases (say there are $k$ of them), then $n$ is “probably prime” with an error probability of at most $0.25^k$.

To prove the integer $n$ is composite, we look for a base $a$ for which $n$ is not a strong probable prime. Such a value of $a$ is also called a Miller-Rabin witness for the compositeness of $n$. For primality, the Miller-Rabin test does not give a mathematical proof that a number is prime. The Miller-Rabin test is a probable prime test. It gives strong evidence that $n$ is a prime number, with an error probability that can be made arbitrarily small by using a large random sample of values of $a$.

Take the prime candidate $P_{297}$ that is discussed above. We plan to run the Miller-Rabin test on $P_{297}$ using 40 random values of $a$ where $1. If $P_{297}$ is shown to be a strong probable prime in all 40 bases, then the prime candidate $P_{297}$ is likely a prime number with an error probability of at most $0.25^{40}$. This probability works out to be less than 1 in 10 raised to 24 (hence the one in a septillion that is mentioned earlier). If one wants stronger evidence, we can compute for more values of $a$. Thus if $P_{297}$ is in actuality a composite number, there is at most a one in septillion chance that the Miller-Rabin test will declare $P_{297}$ is a prime number.

How can the Miller-Rabin test make the claim of having such a small error probability? The fact the the error probability of Miller-Rabin can be made arbitrarily small stems from the following fact.

Theorem 2
Suppose that $n$ is a composite odd number. At most 25% of the numbers in the interval $1 are bases in which $n$ is a strong pseudoprime. Putting it in another way, at least 75% of the numbers in $1 are bases in which $n$ is not a strong pseudoprime.

To paraphrase Theorem 2, if $n$ is composite to begin with, at least 75% of the numbers in $1 will prove its compositeness. That means that at most 25% of the numbers $a$ will exhibit the prime-like property described in Theorem 1. The power of Miller-Rabin comes from the fact that for composite numbers there are more values of $a$ that will give a correct result (in fact, at least 3 times more).

Thus if you apply the Miller-Rabin test on a composite number $n$, you will bound to stumble on a base $a$ that will prove its compositeness, especially if the bases are randomly chosen. Any random choice of $a$ where $1 has at least a 75% chance of being correct on the composite number $n$. In a series of 100 random choices of $a$, it will be hard to miss such values of $a$. The only way that Miller-Rabin can make a mistake by declaring a composite number as prime is to pick all the values of $a$ from the (at most) 25% of the pool of values of $a$ that are strong pseudoprime prime. This probability is bounded by $0.25^k$ (if $k$ is the number of selections of $a$).

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Applying Miller-Rabin on the prime candidate

The first task is to factor $P_{297}-1$. We find that $P_{297}-1=2^3 \times q$ where $q$ is the following odd number:

$q=$
14474011154664524427946373126085988481658748083205
070504932198000989141205029

For each randomly selected $a$, we calculate the following sequence:

$a^q, \ a^{2q}, \ a^{4q}, \ a^{8q} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$

The first term is calculated using the fast powering algorithm (a series of squarings and multiplications). Each subsequent term is the square of the preceding term. Each term in the sequence is reduced modulo $P_{297}$. The goal is to see if the two conditions in Theorem 1 are satisfied. One is that one of the 4 values in (2) is a 1. The other is that the term preceding the first 1 in (2) has to be a -1.

The following shows the 40 numbers that are randomly chosen in the interval $1.

$a_1=$
03006957708701194503849170682264647623506815369915
7798209693214442533348380872
$a_2=$
02223067440101780765895379553626469438082041828085
0568523022714143509352911267
$a_3=$
04531895131849504635258523281146698909008537921009
6337435091877410129499153591
$a_4=$
05434508269932993379745836263818598804800824522102
0278113825689716192402178622
$a_5=$
08799241442673378780142202326330306495270149563840
3866810486309815815031353521
$a_6=$
02638607393577034288802880492058261281940769238659
8928068666401909247319838064
$a_7=$
04283430251977183138176255955338099404217762991191
9192783003754562986178981473
$a_8=$
09773398144692973692102006868849010147546139698798
3443958657834362269077067224
$a_9=$
05504666974469005713839308880951115507992521746498
7157086751623602877205126361
$a_{10}=$
11369425784373951812019794994427515082375862595853
6524984616385315102874812557
$a_{11}=$
11280428157869817083329641054154150272024966029283
2165114734540900026838117128
$a_{12}=$
11208322317253928483879618989535357346499197200982
7728283667193655956607063861
$a_{13}=$
05585951853297694372636067012444311272073854408338
4421611399136081624631900538
$a_{14}=$
06831924581003106427566658433259804779354874917795
9811865334330929987281859876
$a_{15}=$
07339174229323952008915772840377019251465052264221
1294344032116313026124007734
$a_{16}=$
05117387267263929174559713098463717229625661656017
7194611080485470890280573816
$a_{17}=$
06599941646668915168578091934085890873056463577356
8090503454939353325803291530
$a_{18}=$
07545265152740184887140788322673806569482388835389
5577110370797470603035554930
$a_{19}=$
02591621894664804222839429868664505564743756550515
2520842332602724614579447809
$a_{20}=$
04791002227899384351266879075743764807247161403811
8767378458621521760044966007
$a_{21}=$
03251071871924939761772100645669847224066002842238
6690935371046248267119967874
$a_{22}=$
07211128555514235391448579740428274673170438137060
9390617781010839144521896079
$a_{23}=$
02839820419745979344283855308465698534375525126267
1701870835230228506944995955
$a_{24}=$
06304631891686637702274634195264042846471748931602
4893381338158934204519928855
$a_{25}=$
06492095235781034422561843267711627481401158404402
2978856782776323231230432687
$a_{26}=$
11078868891712009912929762366314190797941038596568
5459274315695355251764942151
$a_{27}=$
05795069944009506186885816367149671702413127414386
2708093175566185349033983346
$a_{28}=$
01712922833914010148104423892201355622294341143990
7524285008693345292476544524
$a_{29}=$
09743541325262594740093734822046739122734773994479
9814337973200740861495044676
$a_{30}=$
02503872375817370838455279068302037475992008315394
2976462871038003917493744995
$a_{31}=$
06980677383898331402575574511880992071872803011356
6498794763450065008785347168
$a_{32}=$
01507075889390134242331585173319278262699562685820
7121480322563439665642035394
$a_{33}=$
02471785068822350832987019936892052187736451275830
5372059292781558599916131031
$a_{34}=$
10950891460180297156465120507537244257810396062906
9207306297501015755045004254
$a_{35}=$
11052976297188507170707306917942099264941855478856
2965936913589165233381674539
$a_{36}=$
03911878231948499128291863266472008604449261315172
1053813631612297577166335941
$a_{37}=$
06903294587603383022211116535092146484651980588002
9291840261276683214113088012
$a_{38}=$
03942020579038616658412018517396703874933208670283
3087287933190554281896471934
$a_{39}=$
04338728160253711124705740270085271024911573570055
1690460857511205663297661796
$a_{40}=$
06707597137792150532106913489524457238449067437061
7211249957355483821516113140

For each random number $a_j$, we calculated the 4 numbers indicated in sequence (2). The following 3 tables show the results of the calculation.

$\left[\begin{array}{rrrrrrrrr} j & \text{ } & a_j^q & \text{ } & a_j^{2q} & \text{ } & a_j^{4q} & \text{ } & a_j^{8q} \\ \text{ } & \text{ } & \text{ } \\ 1 & \text{ } & \text{ } & \text{ } & -1 & \text{ } & 1 & \text{ } & 1 \\ 2 & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & -1 & \text{ } & 1 \\ 3 & \text{ } & 1 & \text{ } & 1 & \text{ } & 1 & \text{ } & 1 \\ 4 & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & -1 & \text{ } & 1 \\ 5 & \text{ } & -1 & \text{ } & 1 & \text{ } & 1 & \text{ } & 1 \\ 6 & \text{ } & \text{ } & \text{ } & -1 & \text{ } & 1 & \text{ } & 1 \\ 7 & \text{ } & \text{ } & \text{ } & -1 & \text{ } & 1 & \text{ } & 1 \\ 8 & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & -1 & \text{ } & 1 \\ 9 & \text{ } & \text{ } & \text{ } & -1 & \text{ } & 1 & \text{ } & 1 \\ 10 & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & -1 & \text{ } & 1 \\ 11 & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & -1 & \text{ } & 1 \\ 12 & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & -1 & \text{ } & 1 \\ 13 & \text{ } & -1 & \text{ } & 1 & \text{ } & 1 & \text{ } & 1 \\ 14 & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & -1 & \text{ } & 1 \\ 15 & \text{ } & \text{ } & \text{ } & -1 & \text{ } & 1 & \text{ } & 1 \end{array}\right]$

$\left[\begin{array}{rrrrrrrrr} j & \text{ } & a_j^q & \text{ } & a_j^{2q} & \text{ } & a_j^{4q} & \text{ } & a_j^{8q} \\ \text{ } & \text{ } & \text{ } \\ 16 & \text{ } & 1 & \text{ } & 1 & \text{ } & 1 & \text{ } & 1 \\ 17 & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & -1 & \text{ } & 1 \\ 18 & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & -1 & \text{ } & 1 \\ 19 & \text{ } & \text{ } & \text{ } & -1 & \text{ } & 1 & \text{ } & 1 \\ 20 & \text{ } & 1 & \text{ } & 1 & \text{ } & 1 & \text{ } & 1 \\ 21 & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & -1 & \text{ } & 1 \\ 22 & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & -1 & \text{ } & 1 \\ 23 & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & -1 & \text{ } & 1 \\ 24 & \text{ } & 1 & \text{ } & 1 & \text{ } & 1 & \text{ } & 1 \\ 25 & \text{ } & \text{ } & \text{ } & -1 & \text{ } & 1 & \text{ } & 1 \\ 26 & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & -1 & \text{ } & 1 \\ 27 & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & -1 & \text{ } & 1 \\ 28 & \text{ } & 1 & \text{ } & 1 & \text{ } & 1 & \text{ } & 1 \\ 29 & \text{ } & 1 & \text{ } & 1 & \text{ } & 1 & \text{ } & 1 \\ 30 & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & -1 & \text{ } & 1 \end{array}\right]$

$\left[\begin{array}{rrrrrrrrr} j & \text{ } & a_j^q & \text{ } & a_j^{2q} & \text{ } & a_j^{4q} & \text{ } & a_j^{8q} \\ \text{ } & \text{ } & \text{ } \\ 31 & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & -1 & \text{ } & 1 \\ 32 & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & -1 & \text{ } & 1 \\ 33 & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & -1 & \text{ } & 1 \\ 34 & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & -1 & \text{ } & 1 \\ 35 & \text{ } & 1 & \text{ } & 1 & \text{ } & 1 & \text{ } & 1 \\ 36 & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & -1 & \text{ } & 1 \\ 37 & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & -1 & \text{ } & 1 \\ 38 & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & -1 & \text{ } & 1 \\ 39 & \text{ } & 1 & \text{ } & 1 & \text{ } & 1 & \text{ } & 1 \\ 40 & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & -1 & \text{ } & 1 \end{array}\right]$

There are 4 columns of calculation results, one for each term in sequence (2). If a calculation result is a blank in the above tables, it means that the result is a number that is not 1 or -1 modulo $P_{297}$. For example, $a_1^q \ (\text{mod} \ P_{297})$ and $a_2^q \ (\text{mod} \ P_{297})$ are congruent to the following two numbers:

$a_1^q \equiv$
86168678768024029811437552745042076645410792873480
629834883948094184848812907

$a_2^q \equiv$
10235477176842589582260882228891913141693105976929
7597880545619812030150151760

In the above 3 tables, all results match the conditions of Theorem 1. For each number $a_j$, the calculated results are eventually 1. On some of the rows, the first result is a 1. In all the other rows, the term right before the first 1 is a -1. For example, in the first row where $j=1$, the first 1 $a_1^{4q}$ and the term preceding that is a -1.

The results in the above 3 tables show that the number $P_{297}$ is a strong probable prime in all 40 of the randomly chosen bases. We have very strong evidence that the number $P_{297}$ is a prime number. The probability that it is a composite number but we mistakenly identify it as prime is at most one in a septillion!

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Exercise

In our search for probable primes larger than the 8th Fermat number, we also find that the number $P_{301}=2^{301}+301$ is also a probable prime base 2. The following shows the decimal digits:

$P_{301}=$
11579208923731619542357098500868790785326998466564
0564039457584007913129640237

Is it a prime number? Perform the Miller-Rabin test on this number.

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$\copyright \ \ 2014 \ \text{Dan Ma}$

# Is factorization a hard problem?

Is factorization a hard problem? There is plenty of empirical evidence that it is so. Take the following 309-digit number that is known as RSA-1024, an example of an RSA number.

RSA-1024
13506641086599522334960321627880596993888147560566
70275244851438515265106048595338339402871505719094
41798207282164471551373680419703964191743046496589
27425623934102086438320211037295872576235850964311
05640735015081875106765946292055636855294752135008
52879416377328533906109750544334999811150056977236
890927563

RSA-1024, a 1024-bit number, is a product of two prime numbers $p$ and $q$. No one has been able to factor this number, despite the advances in factoring algorithms and computing technology in recent decades. RSA-1024 is part of the RSA Factoring Challenge that was created in 1991. Even though the challenge was withdrawn in 2007, it is believed that people are still taking up the challenge to factor this and other unfactored RSA numbers. In fact, the successful factoring of RSA-1024 or similarly sized numbers would have huge security implication for the RSA algorithm. The RSA cryptosystem is built on the difficulty (if not the impossibility) of factoring large numbers such as RSA-1024.

Yet it is very easy to demonstrate that RSA-1024 is not a prime number. The fact that it is composite can be settled by performing one modular exponentiation. Denote RSA-1024 by $N$. We compute $2^{N-1} \ (\text{mod} \ N)$.

We find that $2^{N-1} \equiv T \ (\text{mod} \ N)$ where $T$ is the following 309-digit number.

$T=$
12093909443203361586765059535295699686754009846358
89512389028083675567339322020593385334853414711666
28419681241072885123739040710771394053528488357104
98409193003137847878952260296151232848795137981274
06300472693925500331497519103479951096634123177725
21248297950196643140069546889855131459759160570963
857373851

Obviously $T$ is not 1. This fact is enough to prove that the modulus $N$ is not a prime number. This is because the number $N$ lacks a property possessed by all prime numbers. According to Fermat’s little theorem, if $N$ were prime, then that $a^{N-1} \equiv 1 \ (\text{mod} \ N)$ for all integers $a$ that are relatively prime to $N$. In particular, if $N$ were prime, then we would have $2^{N-1} \equiv 1 \ (\text{mod} \ N)$, the opposite of our result.

The modular exponentiation $a^{N-1} \ (\text{mod} \ N)$ discussed here can be performed using the fast powering algorithm, which runs in polynomial time. In the fast powering algorithm, the binary expansion of the exponent is used to convert the modular exponentiation into a series of squarings and multiplications. If the exponent $N-1$ is a $k$-bit number, then it takes $k-1$ squarings and at most $k-1$ multiplications. For RSA-1024, it takes 1023 squarings and at most 1023 multiplications (in this instance exactly 507 multiplications). This calculation, implemented in a modern computer, can be done in seconds.

The above calculation is a vivid demonstration that factoring is hard while detecting the primality or compositeness of a number is a much simpler problem.

The minimum RSA key length prior to the end of 2013 is 1024. After 2013, The minimum RSA key length is 2048. In fact, the largest RSA number is RSA-2048 (has 2048 bits and 617 decimal digits), which is expected to stay unfactored for years to come barring dramatic advances in factoring algorithms or computing capabilities.

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Exercise

Using a software package that can handle modular exponentiation involving large numbers, it is easy to check for “prime versus composite” status of a large number. Find a number $n$ whose prime factorization is not known. Either use known numbers such as RSA numbers or randomly generate a large number. Then calculate the modular exponentiation $a^{n-1} \ (\text{mod} \ a)$ for several values of $a$ (it is a good practice to start with $a=2$). If the answer is not congruent to 1 for one value of $a$, then we know $n$ is composite. If the exponentiation is all congruent to 1 for the several values of $a$, then $n$ is a likely a prime number.

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$\copyright \ \ 2014 \ \text{Dan Ma}$

# A primality testing exercise from the 9th Fermat number

In this post, we discuss a primality testing exercise on a number derived from a prime factor of the 9th Fermat number. The idea of the exercise is that we test the primality of a number generated from a known large prime number by tagging or inserting a digit into the known prime number. This is an excellent way to generate exercises for primality testing. This approach of altering a known prime number is not meant to be a prime number generation process.

A Fermat number is a positive integer of the form $F_n=2^{2^n}+1$ where $n=0,1,2,\cdots$. The 9th Fermat number is $F_9=2^{2^9}+1=2^{512}+1$. This is a 155-digit number that was completely factored in 1993 [1]. Its three prime factors have 7 digits, 49 digits and 99 digits. The following is the 49-digit factor of $F_9$:

7,455,602,825,647,884,208,337,395,736,200,454,918,783,366,342,657

The above 49-digit number is a prime number that was resulted from a published effort of factorizing $F_9$. The other two factors of $F_9$ can be found in [1] or here. If we tag a digit in front of this 49-digit number, say the digit 6, is the resulting number a prime number? In this post, we discuss the primality testing of the following 50-digit number:

$A=$ 67,455,602,825,647,884,208,337,395,736,200,454,918,783,366,342,657

Primality testing is an important aspect of public key cryptography. For example, the modulus in the public key in RSA cryptosystem is a product of two distinct large prime numbers. To generate the key, it is crucial to generate large numbers at random and to be able to efficiently test whether the numbers are prime.

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Pre-checks before primality testing

Primality tests such as Fermat test and Miller-Rabin test are popular and are easy to use. These tests check for the primality of a given integer $n$ by randomly choosing a set of auxiliary numbers and then checking whether any one of these auxiliary values is a witness to the compositeness of the number $n$. We would like to emphasize the point that before applying a randomized primality testing procedure, it will be advantageous to perform some easy checks to quickly rule out primality. If these quick checks show that the number being tested is composite, we can then continue the search by moving on to another candidate prime number. With the above number $A$ as example, we perform the following pre-checks.

• Check small prime numbers just in case the number being tested has small prime factors.
• If the previous step produces no factors, look for small Fermat witnesses.

The idea of the first check is clear. A Fermat witness for an integer $n$ is an integer $a$ that is relatively prime to $n$ such that $a^{n-1} \not \equiv 1 \ (\text{mod} \ n)$. If we can find such a number $a$, we can say conclusively that $n$ is not a prime number. According to Fermat’s little theorem, if $n$ is a prime number, then the congruence would be a one, i.e., $a^{n-1} \equiv 1 \ (\text{mod} \ n)$. This idea is the basis for the Fermat primality test, which is performed on a random set of values of $a$. Here in the pre-check, we calculate $a^{n-1} \ (\text{mod} \ n)$ for some small values of $a$ such as 2 or 3. The idea is to quickly rule out the candidate prime numbers that happen to have a small Fermat witness.

If either one of the two checks indicate the number being tested is composite, then we pick another number to test. If both of the above steps fail to indicate compositeness, we can then proceed to use the “probabilistic” part of the Fermat test or another probabilistic primality test.

We like to point out that at the outset, the author of the blog did not know the primality status of the 50-digit number $A$ that is obtained by tagging a 6 in front of the 49-digit factor of the 9th Fermat number. Though this number is far too small to work in the RSA algorithm, it serves as an excellent exercise for primality testing.

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Pre-checking the number $A$

Very quickly, 7 is found to be a factor of $A$. We have $A=7 \times B$ where $B$ is:

$B=$ 9636514689378269172619627962314350702683338048951

We then try to find small factors of $B$. The number 283 is found to be a factor of $B$. Then we have $A=7 \times 283 \times C$ where $C$ is the following 47-digit number:

$C=$ 34051288655046887535758402693690285168492360597

By tagging a 6 in front of the 49-digit factor of the 9th Fermat number, we obtain a composite number. The prime or composite status for the number $A$ is settled. Now the next question: is the remaining factor $C$ prime or composite?

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Pre-checking the number $C$

We perform the two pre-checks as described above. In searching for small factors, we find that $C$ is not divisible by all prime numbers less than 200,000. So if $C$ has a factor, it will have to be a large one. Or it could be that $C$ is prime. Instead of continuing to search for factors, we can do the second check. We calculate $2^{C-1} \ (\text{mod} \ C)$ and find that:

$2^{C-1} \equiv t \ (\text{mod} \ C)$

where $t=$ 7426390354305013563302537374271875139618265902

Since $t$ clearly is not 1, this tells us that $C$ is composite.

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Remarks

Instead of doing the randomized primality testing right away, it makes sense to do the pre-checks as described above so that we do not waste time and effort on composite numbers that have small factors or have small Fermat witnesses. As soon as these two checks indicate the compositeness of a number, we then move on to another candidate prime number. If a number passes these two checks, then we can apply a primality test such as the Fermat test or the Miller-Rabin test.

Another remark is that primality testing is easy while factoring is hard. The compositeness of the 47-digit number $C$ can be determined by just one modular exponentiation calculation, which can be computed using the fast powering algorithm. This algorithm uses the binary expansion of the exponent to convert the modular exponentiation into a series of squarings and multiplications. Because the exponent $C-1$ is a 155-bit number, computing $2^{C-1} \ (\text{mod} \ C)$ only requires 154 squarings and at most 154 multiplications (the actual count of multiplications is 76). Implemented in a modern computer, this calculation takes less than one second.

On the other hand, factoring the above 47-digit number $C$ by the brute force approach (checking every possible prime factor) would require checking approximately $2 \times 10^{20}$ many prime numbers as potential factors. The age of the universe is believed to be about 13 billions years, which is about $4 \times 10^{17}$ seconds. That’s why we stop checking for prime factors of $C$ at some point. The authors of [1] factored a 155-digit number in 1993. So the current computing technology is certainly capable of factoring a 47-digit number. But the effort will take time and will require using more advanced factoring algorithm.

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Exercises

The idea of the exercise is that we test the primality of a number generated from a known large prime number by tagging or inserting a digit into the known prime number. Using the same 49-digit factor of $F_9$, we can tag another digit in front or insert a digit somewhere in the middle. Another possibility is to try the other two factors of $F_9$.

There are many other large primes to be found on the Internet that can be used for this exercise. For example, factors of other Fermat numbers and RSA numbers.

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Reference

1. Lenstra A. K., Lenstra H. W., Manasse M. S., Pollard J. M. The Factorization of the Ninth Fermat Number, Mathematiics of Computation, Volume 61, Number 203, July 1993, Pages 319-349.

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$\copyright \ \ 2014 \ \text{Dan Ma}$