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For all $a,b \in \mathbb{Z}$ and for all $m,n \in \mathbb{N}\setminus \left\lbrace0\right\rbrace$,

is $a^{48m+1}+b^{48n+1} \equiv 0 \pmod{39} \iff a+b \equiv 0 \pmod{39}$?

I think the answer is yes, but I can't prove it. Is there somebody who can help me? Thank you.

Bill Dubuque
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Nguyen Thy
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3 Answers3

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Hint: Use Euler theorem $$a^{\varphi(n)}\equiv 1\pmod n$$ if $\gcd(a,n )= 1$. Notice that $$\varphi (39) = 2\cdot 12 =24$$

Edit: 12. 29. 2019

  • a) If $(a,39) =1$ then clearly also $(b,39)=1$ so we have $a^{24} \equiv 1$ and $b^{24} \equiv 1$ so $a^{48m}a \equiv a$ and $b^{48n}b \equiv b$ and thus: $$a^{48m+1}+b^{48n+1} \equiv a+b$$ and the equivalence follows.

  • b) If $(a,39) \ne 1$ then $3\mid a$ or $13 \mid a$ or $39\mid a$.

    • If $39\mid a$ then also $39\mid b$ and claim follows

    • If $3\mid a$ and $13\nmid a$ then if $3\mid a^{48m+1}+b^{48n+1} $ we have also $3\mid b$ so $3\mid a+b$ and vice versa. And for $13$ you can repeat a process in case a).

    • If $13\mid a$ and $3\nmid a$ then repeat a process from previous case

nonuser
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Hint

Using http://mathworld.wolfram.com/FermatsLittleTheorem.html

For $3\mid x(x^2-1),$

and $13|x(x^{12}-1)$

$[3,13]$ will divide $x(x^{[2,12]}-1)$ and hence $x(x^{12m}-1)$ for any integer $m$

lab bhattacharjee
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  • @Downvoter, please find the updated answer – lab bhattacharjee Dec 28 '19 at 17:23
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    I didn't down voted but the hint can be elaborated a little more so that the O.P can understand. – The Demonix _ Hermit Dec 28 '19 at 17:24
  • Can you explain more clearly why $13|x(x^{12}-1)$? – Nguyen Thy Dec 28 '19 at 17:32
  • @Nguyen, see http://mathworld.wolfram.com/FermatsLittleTheorem.html and the updated post – lab bhattacharjee Dec 28 '19 at 17:34
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    Either $\gcd(x, 13) = 1$ and so $x^{12}\equiv 1 \pmod {13}$ and $13|x^{12} -1$. or $\gcd(x,13) = 13$ and so $13|x$. Either way, $13|x\cdot (x^{12}-1)$.... Another way of saying this is if $p$ is prime then $a^p \equiv a$ (whether $a$ is or is not a multiple of $p$. [If $a$ isn't a multiple of $p$ then $a^{p-1}\equiv 1$ so $a^{p-1}a\equiv a$. And if $a$ is a multiple of $p$ then $a \equiv 0$ and $a^p \equiv 0$.]) – fleablood Dec 28 '19 at 18:05
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Note that $39=3\times13$, and $3$ and $13 $ are prime.

Using Fermat's little theorem, $a^3\equiv a\bmod 3$,

and by induction $a^{2k+1}\equiv a\bmod 3$, so $a^{48m+1}\equiv a\bmod 3$.

Likewise, $a^{13}\equiv a\bmod 13,$ so $a^{12n+1}\equiv a\bmod 13$, so $a^{48m+1}\equiv a\bmod 13$.

Therefore, $a^{48 m+1}\equiv a\bmod 39$. Can you take it from here?

J. W. Tanner
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  • Can you explain more why $a^{13}\equiv a \pmod{13} $ then $a^{12n+1} \equiv a \pmod{13} $ and why $a^{12n+1} \equiv a \pmod{13} $ then $a^{48m+1} \equiv a \pmod{13}$? Thank you. – Nguyen Thy Dec 29 '19 at 05:41
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    to answer the second part of your comment, take $n=4m$, so $48m+1=12(4m)+1$ – J. W. Tanner Dec 29 '19 at 05:45
  • Ok I see, thank you. – Nguyen Thy Dec 29 '19 at 05:48
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    for the first part, do you know mathematical induction? base case $n=1$: $a^{13}\equiv a$, now if $a^{12(n-1)+1}\equiv a$ then $a^{12n+1}=a^{13}a^{12(n-1)}\equiv aa^{12n-12}\equiv a^{12n-11}\equiv a^{12(n-1)+1}\equiv a$ – J. W. Tanner Dec 29 '19 at 05:53