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Problem: Let $m$ and $n$ be positive integers such that $\gcd(m,n)=1.$Compute $\gcd(5^m+7^m,5^n+7^n)$.

My Attempt: I wrote a simple computer program and deduced that $\gcd(5^m+7^m,5^n+7^n)=2$ if $m+n$ is odd and $\gcd(5^m+7^m,5^n+7^n)=12$ if $m+n$ is even. I would like to know if there is a way to prove these results.

Edit: Found a solution on the web:enter image description here

I don't understand why $\gcd(s_m,s_n)=\gcd(s_m,s_{n-2m})$. According to me, $\gcd(s_m,s_n)=\gcd(s_m,5^m7^ms_{n-2m})$.

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  • $\begingroup$ Well 7-5 =2 and 7+5 = 12 is just too good a coincidence to ignore. Gcd (5^m+7^m,5^n+7^n)=gcd (5^m+7^m-5^n-7^n,5^n+7^n). How can we rewrite $5^m-5^n+7^m-7n $? $\endgroup$
    – fleablood
    Nov 4, 2016 at 7:27
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    $\begingroup$ You are correct that gcd (sm,sn)=gcd (sm,5^m7^msn-2n). But as neither 5 nor 7 divide sm, we can divide out all factors of 5 and 7. $\endgroup$
    – fleablood
    Nov 4, 2016 at 9:29
  • $\begingroup$ That's basically my answer. Go throough my example. See where I divide 5 out of the gcd. $\endgroup$
    – fleablood
    Nov 4, 2016 at 9:32
  • $\begingroup$ If gcd (n,b)=1 then gcd (b,nc)=gcd (b,c). You can verify gcd (7,5^n+7^n)=1 so gcd (5^m7^m,sm)=1. So gcd (sm,5^m7^msn-2m)=gcd (sm,sn-2m). $\endgroup$
    – fleablood
    Nov 4, 2016 at 9:41

1 Answer 1

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Here's a simple case. See if you can generalize it:

$\gcd (5^3+7^3,5^7+7^7)=$

$\gcd (5^3+7^3,5^7+7^7-7^4 (5^3+7^3))=$

$\gcd (5^3+7^3,5^7-5^3*7^4)=$ note: $5\not \mid 5^3+7^3$ so

$\gcd (5^3+7^3,5^4-7^4)=$

$\gcd (5^3+7^3,5^4-7^4+7 (5^3+7^3))=$

$\gcd (5^3+7^3,5^4+5^3*7)=$

$\gcd (5^3+7^3,5+7)=$

$\gcd (5^3+7^3-7^2 (5+7),5+7)=$

$\gcd(5^3-7^2*5,5+7)=$

$\gcd(5^2-7^2,5+7)=$

$\gcd (5^2-7^2+7 (5+7),5+7)=$

$\gcd (5^2+7*5,5+7)=$

$\gcd (5+7,5+7)=12$

As $\gcd(m,n)=1$

This "swinging" by reducing and subtracting the powers will eventually result in $5^{\gcd (m,n)}\pm 7^{\gcd (m,n)}=5\pm 7= 2|12$. (Positive/negative doesn't matter for gcd)

Whether it will be $5+7$ or $5-7$ will depend on whether we swing an even or odd number of times.

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Alright. Here's a more thorough proof:

Let $\gcd(B,A) = 1$ and wolog $B > A$ and $\gcd(n,m)=1$ wolog $n>m$ then $\gcd(B^n+A^n, B^m+A^m) = B + (-1)^{m+n}A$ (and so $\gcd(7^n+5^n,7^m+5^m) = 2$ if $m+n$ is odd and is $12$ if $m+n$ is even.

Proof:

$\gcd(n,m)=1$ which means by Euclid's algorithm That there are series of $n_i, m_i;k_i$ so that:

$n=n_0; m= m_0$

$n_0 = k_0*m_0 + m_1$

$n_1 = m_0; n_1 = k_1*m_1 + m_2$

.....

$n_i = m_{i-1}; n_i = k_i*m_i + m_{i+1}$

...

$n_{z-1} = k_{z-1}m_{z-1} + 1; m_z = 1$

$n_z=m_{z-1} = 1*m_{z- 1}$

$n_{z+1} = 1 $

Interestingly $z$ is odd if and only if $n+m$ is odd. (To be honest I'm not sure how to prove that.)

$\gcd(B^n + A^n,B^m + A^m) = \gcd(B^n + A^n - B^{n-m}(B^m + A^m),B^m + A^m)=$

$\gcd(B^n + A^n - B^n - B^{n-m}A^m,B^m + A^m) = \gcd(A^n - A^mB^{n-m},B^m +A^m)=$

$\gcd(-A^m(B^{n-m} - A^{n-m}),B^m + A^m) = \gcd(B^{n-m} - A^{n-m},B^m+ A^m)$

And repeating via induction we can get:

$=\gcd(B^{n-k_0*m=m_1} \pm A^{m_1},B^{m=n_1} + A^{n_1})$ where it is a "plus" if $k_0$ is even and a "minus" if $k_0$ is odd.

An repeating via induction we can get:

$=\gcd(B \pm A, B \pm A) = B\pm A$

"plus" if $\sum k_i$ is even, and "minus" if the sum is odd.

Somehow we can show $\sum k_i \text{ is odd } \iff z \text { is even } \iff $m+n $ \text { is even }$. I guess that part needs a little work.

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  • $\begingroup$ For a formal proof I guess we could use induction. $\endgroup$
    – fleablood
    Nov 4, 2016 at 8:47
  • $\begingroup$ I have made an edit. Please refer to it. $\endgroup$
    – Student
    Nov 4, 2016 at 8:50

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