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I am working through a worksheet talking about some of Euler's results and how $X = a^{2^n}+1$ is not necessarily prime.

One example that Euler gave was that $X = 6^{128}+1$ is not prime since it is divisible by $Y = 257$.
How would Euler have gone about showing this result?

I am relatively new to number theory, I can see that $Y = 2*128+1 = 257$ but I do not know if that information is relevant or not.

Bill Dubuque
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Conor_Meise
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  • You could just compute $6^{128} \bmod 257$ by hand, for example. – Misha Lavrov Sep 06 '24 at 14:54
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    Short answer: $6^4$ has remainder $11$ when divided by $257$, or rather $6^4 = a \cdot 257 + 11$ for an $a$ that I don't compute. Then $6^{8} = (6^4)^2 = (a \cdot 257 + 11)^2 = b \cdot 257 + 121$ for an integer $b$ that I don't compute; hence $6^8$ has remainder $121$ after division by $257$. Continuing, $6^{16}, 6^{32}, 6^{64}$, and $6^{128}$ have remainders $249, 64, 241$, and $256$. This is all computed by hand. Thus $6^{128} = c \cdot 257 + 256$, and finally $6^{128} + 1 = (c + 1) \cdot 257$. We didn't find $c$ (or $b$ or $a$). This is easier to say in the language of modular arithmetic. – davidlowryduda Sep 06 '24 at 15:02
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    By fermat's little theorem if $a$ is not a multiple of $257$ then $a^{256}\equiv 1 \pmod {257}$ so $a^{128}\equiv \pm 1\pmod{257}$. It's a matter of finding an $a$ where $a^{128}\equiv -1$. Now euler was probably smarter than I am this morning so he probably had a better way but I'd try successive squaring on $2$ and/or $3$ and it'd be easy to get a result. – fleablood Sep 06 '24 at 15:04
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    I do not think it is a valid Dup , @lulu , though it may eventually help OP here. – Prem Sep 06 '24 at 15:04
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    You could just compute that 6 is not a quadratic residue (mod 257), therefore $6^{(257-1)/2} \equiv -1 \pmod{257}$. – Daniel Schepler Sep 06 '24 at 15:06
  • @Prem Yes, I agree. For whatever reason, I mentally changed the $6$ into a $2$. Have deleted the duplicate vote. – lulu Sep 06 '24 at 15:06
  • In case this information is at all helpful, the number in question is divisible by 257, 763649, 50307329, 3191106049, and maybe by something else that I don't have the computational resources to check. – Dan Sep 06 '24 at 15:15
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    The Question here is not to show that $257$ works. It is to know how Euler got that number. When we know that number , we could use many techniques ( listed the comments ! ) to show that it works. – Prem Sep 06 '24 at 15:18
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    @Prem I agree that's the more interesting thing to ask, but that doesn't seem to be what the text of the question is actually asking. It's asking "how would Euler have gone about showing [that $6^{128}+1$ is divisible by $257$]?" not where Euler would have gotten those numbers to begin with. – Misha Lavrov Sep 06 '24 at 15:30
  • I think those two issues are intertwined : Either Euler "knew" some general theorem , @MishaLavrov , or Euler "knew" why $257$ will work here , even before trying to generate a Proof of that fact. – Prem Sep 06 '24 at 15:53
  • Special case of Euler's Criterion. $\ \ $ – Bill Dubuque Sep 06 '24 at 20:49

2 Answers2

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Euler has given the answer himself via the Paper here :

"Observations on a theorem of Fermat and others on looking at prime numbers"

At the Conclusion , Euler claims :

"Theorem V. $3^n + 2^n$ is able to be divided by $2n + 1$ if $n$ is equal to either $12p + 3$, $12p + 5$, $12p + 6$ or $12p + 8$. Moreover, $3^n − 2^n$ is able to be divided by $2n + 1$ if $n$ is equal to either $12p$, $12p + 2$, $12p + 9$ or $12p + 11$."

"Theorem VI. Under these very same conditions for $3^n + 2^n$ , $6^n + 1$ is also able to be divided by $2n + 1$ , and $6n − 1$ too under those for $3n − 2n$."

Here , $128 = 12 \times 10 + 8$ , hence it will satisfy these two theorems.
We can take $n = 128$ to get $6^{128} + 1$ which will have the Divisor $2 \times 128 + 1 = 257$

Details :

https://web.math.ucsb.edu/~agboola/teaching/2005/winter/old-115A/euler.pdf

Prem
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$Y=257=2^8+1\mid(2^8+1)(2^8-1)(2^{16}+1)(2^{32}+1)(2^{64}+1)=2^{128}-1$,

so $2^{128}\equiv1\pmod Y$, so $\left(\dfrac2{257}\right)=1$ by Euler's criterion.

(Alternatively, $\left(\dfrac2{257}\right)=1$ because $Y\equiv1\pmod8$.)

By quadratic reciprocity,

$\left(\dfrac3{257}\right)\left(\dfrac{257}{3}\right)=(-1)^{\dfrac{3-1}{2}\dfrac{257-1}{2}}$, so $\left(\dfrac3{257}\right)\left(\dfrac{2}{3}\right)=1,$ so $\left(\dfrac3{257}\right)=-1$.

(Alternatively, $\left(\dfrac3{257}\right) =-1$ because $Y\equiv5\pmod{12}$.)

In any event, it follows that $\left(\dfrac6{257}\right)=-1$, so by Euler's criterion $6^{128}\equiv-1\pmod Y$;

i.e., $Y\mid X$.

J. W. Tanner
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    The question does not ask for any proof (which is a duplicate). Rather it asks how Euler proved it. So unless you have historical documents supporting the above then it is not an answer. – Bill Dubuque Sep 07 '24 at 20:17
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    Didn’t Euler know quadratic reciprocity and Euler’s criterion? – J. W. Tanner Sep 08 '24 at 12:49
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    @BillDubuque Strictly speaking, your comment is accurate. However, I feel that a reasonable interpretation of the posted question is : how might Euler have proven it. – user2661923 Sep 10 '24 at 22:39
  • @user2661923 If not interpreted as a purely historical question then it is a dupe of a FAQ (and I already mentioned Euler's Criterion a day prior to the OP). Generally vague questions often provide means for rampant duplication - so they need to be carefully managed. – Bill Dubuque Sep 10 '24 at 23:23