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I need to prove, using Bézout's identity, that $5^n+6^n$ and $5^{n+1}+6^{n+1}$ are coprime for all $n \in \mathbb{N}^*$. I know that if they are coprime there exist $u,v \in \mathbb{Z}$ such that:

$u(5^n+6^n)+v(5^{n+1}+6^{n+1})=1$,

but I am unsure on how to proceed from here. Any help would be much appreciated.

Toño Martín Monge
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4 Answers4

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Let $$a_n=5^n+6^n$$

Since $5,6$ are the roots of $$0=(x-5)(x-6)=x^2-11x+30$$ we deduce that the $a_n$ satisfy the recursion $$a_n=11a_{n-1}-30a_{n-2}\quad a_0=2\quad a_1=11$$

From this it is clear that if any $a_n,a_{n-1}$ have a common factor for any $n$, that factor also divides $a_{n-2}$. (Note: we have $\gcd(a_n,30)=1$ for all $n$ so we can disregard the coefficient of the $a_{n-2}$ term). Since $a_0,a_1$ have no common factors, we are done.

lulu
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  • With a little more effort, you can also deduce a Bezout relation, which seems to be more what the OP is looking for. – Ewan Delanoy Feb 19 '19 at 17:05
  • It is clear that a common factor of $a_n,a_{n-1}$ also divides $30a_{n-2}$. How do you get rid of $30$? – lhf Feb 19 '19 at 17:24
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    @lhf It follows from $a_n=5^n+6^n$ that $a_n$ is coprime to $5$ and $6$ – Ewan Delanoy Feb 19 '19 at 17:26
  • @lhf Yes, I should have stressed that we know that no common factor could be a divisor of $30$. I'll edit to point that out. – lulu Feb 19 '19 at 17:30
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You want Bezout's identity? Here is one.

$(6^n,5^n)=1$,so integers $s,t$ exist such that $s*6^n+t*5^n=1$.

Then we have $(s-t)(6^{n+1}+5^{n+1})+(-5s+6t)(6^n+5^n)=1$

Bill Dubuque
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StAKmod
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Hint $ $ By Euclid: $(a\!+\!b,\,5a\!+\!6b) = (a\!+\!b,\,5a\!+\!6b-5(a\!+\!b)) = (a\!+\!b,\,b) = (a,b)$

Bill Dubuque
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  • Oh cool this can be generalized with $m$ and $m+1$ so $(a!+!b,,ma!+!(m!+!1)b) = (a!+!b,,ma!+!(m!+!1)b-m(a!+!b)) = (a!+!b,,b) = (a,b)$

    $m,n,p\in\Bbb{N}|n\neq p,m>1$

    $m^n!+!(m!+!1)^n$ is coprime to $m^p!+!(m!+!1)^p$

     – quantus14 Feb 25 '19 at 18:08
  • @quantus14 In fact it can be generalized further as here. – Bill Dubuque Feb 25 '19 at 18:26
  • @Bill_Dubuque So if $a,b,m,s,t,u,v\in\Bbb{N}|su-tv=\pm 1$ and $s,t,u,v>1$

    then $(sa+tb,ua+vb)=(a,b)\Rightarrow s^m+t^m$ is coprime to $u^m+v^m$?

     – quantus14 Feb 25 '19 at 19:46
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You can also do it as, let $d_n=(5^n+6^n,5^{n+1}+6^{n+1})$. Since $d_n\mid 5^n+6^n$, $d_n\mid 5^{n+1}+5\cdot 6^n$, hence, $d_n\mid 5^{n+1}+6^{n+1}-(5^{n+1}+5\cdot 6^n)=6^n$. If $p\mid d_n$ a prime, then $p\in\{2,3\}$. Now, run the same argument, this time noticing $d_n\mid (6\cdot 5^n+6^{n+1})-(5^{n+1}+6^{n+1})=5^n$, hence, any $p\mid d_n$ satisfy $p\in\{5\}$, contradiction.

TBTD
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