Express $ \operatorname{gcd}\left(5^{m}+7^{m}, 5^{n}+7^{n}\right) $ in terms of $m$ and $n$

Question

Let $m$ be a positive integer with $\operatorname{gcd}(m, n)=1 .$ Express $ \operatorname{gcd}\left(5^{m}+7^{m}, 5^{n}+7^{n}\right) $ in terms of $m$ and $n$

My work -

let $d=\operatorname{gcd}(5^m +7^m,5^n +7^n)$ then

$5^{2m} \equiv 7^{2m}$ mod(d)

$5^{2n} \equiv 7^{2n}$ mod(d)

and obviously $gcd(5,d)=gcd(7,d)=1$ so,

$5^{gcd(2m,2n)} \equiv 7^{gcd(2m,2n)}$ (mod d)

$5^2 \equiv 7^2$ (mod d)

$d= 1,2,3,4,6,8,12,24$

now i find values of d how to express this in terms of $m$ and $n$ ???

A sum $a+b$ divides $a^k+b^k$ if $k$ is odd. Hence, I think that the answer has to do with the odd common divisors of $m$ and $n$. Also, $2$ is obviously a common divisor of $5^m+7^m$ and $5^n+7^n$ as well. — Geoffrey Trang, May 23 '20 at 05:03
Duplicate of https://math.stackexchange.com/questions/262130/how-to-prove-gcdam-bm-an-bn-a-gcdm-n-b-gcdm-n — iam_agf, May 23 '20 at 05:27
Since $,\gcd(5,7)=1,$ the same proof as here works for this homogeneous form (i.e. homogenize that proof). — Bill Dubuque, May 23 '20 at 05:48
@MonsieurGalois Not a dupe of that since this involves sums not differences, so requires further argument. — Bill Dubuque, May 23 '20 at 05:54
@Bill Dubuque yeah, you're right. I deleted it but maybe the order didn't complete. In fact I started to try the problem but the only thing I found is that if m<n and n=mk+r, then d| 5^r-7^r. — iam_agf, May 23 '20 at 06:03
If $m$ and $n$ are both odd. By looking mod 3 and mod 8, you can prove that in this case we have $d=12$. — Julian Mejia, May 23 '20 at 06:16
In the other case, say $m$ even, $n$ odd. We have $3$ does not divide $5^m+7^m$. 4 doesn't divide it either, so $d=2$. — Julian Mejia, May 23 '20 at 06:20

score 2 · Answer 1 · answered May 23 '20 at 06:39

For $k$ odd we have that $$5^k+7^k\equiv 0\mod 3$$ and $$5^k+7^k\equiv 5+7\equiv 4\mod 8$$

For $k$ even we have that $$5^k+7^k\equiv 2\mod 3$$ and $$5^k+7^k\equiv 1+1\equiv 2\mod 8$$

From here we have that if $m,n$ are both odd(i.e. if $m+n$ is even), then $\gcd(5^m+7^m,5^n+7^n)=12$. Otherwise, if between $m$ and $n$ there is an odd and an even number(i.e. if $n+m$ is odd), then $\gcd(5^m+7^m,5^n+7^n)=2$.

So, for $\gcd(m,n)=1$, you could express $$\gcd(5^m+7^m,5^n+7^n)=2\cdot 3^{(m+n+1)\%2}\cdot 2^{(m+n+1)\%2}$$ Where $a\%b$ denotes the remainder that we obtain when dividing $a÷b$.

Ps. This is clearly not a favorable answer since it doesn't seem generalizable.

In fact the general case is not that difficult - I added an answer deriving that. — Bill Dubuque, May 23 '20 at 20:40

Bill Dubuque · Accepted Answer · 2020-05-24T01:53:19.177

Case $\,a,b = 5,7\,$ below [homogenization of this], which applies $\color{#90f}{\rm E} = $ Euclid's algorithm, e.g. $$(a_1,a_2,b) \overset{\color{#90f}{\rm E}}= (\bar a_1,\bar a_2,b)\,\ \ {\rm if}\,\ \ {a_i\equiv \bar a_i}\!\!\! \pmod{\!b},\ \text{is used in the first line of the proof}$$

Theorem $\ $ If $\, m,n\in\Bbb N,\ $ $(m,n)\!=\!1\!=\!(a,b),\,$ and wlog $\,m \!=\! 1\!+\!2j\,$ is odd, then

$$ d := (a^{\large m}\!+\!b^m,a^{\large n}\!+\!b^{\large n})= (a\!+\!b,\color{#0a0}{(-\!1)^{\large n}\!+\!1}) =\begin{cases} (a\!+\!b,2) \ \ {\rm if}\,\ 2\mid n\\ (a\!+\!b)\quad\ \, {\rm if}\ \ 2\nmid \!n\end{cases}\qquad $$

Proof $\ \bmod d\!:\ b^{-1}\,$ exists by $\,(d,b) \overset{\color{#90f}{\rm E}}= (a^m,a^n,b)=1\,$ by $\,(a,b)=1.\,$ Let $\,c \equiv a/b:= ab^{-1}$. Then $\, {c^{\large m}}^{\phantom{|^|}}\!\!\!\equiv -1\equiv c^{\large n}\Rightarrow c^{\large 2m}\equiv 1\equiv c^{\large 2n}$ so $\,{\rm ord}\, c^{\large 2}$ divides coprimes $m,n$ so is $1,\,$ so $\,\color{#c00}{c^{\large 2}\equiv 1}.\,$ $\,{-}1\equiv c^{\large m}\!\equiv c^{\large\phantom{,}}\!(\color{#c00}{c^{\large 2}})^{\large j}\!\equiv c\,$ $\Rightarrow\,c\!+\!1\equiv 0\,\overset{\times\ b}\Rightarrow\,a\!+\!b\equiv 0\,$ so $\,d \overset{\color{#90f}{\rm E}}= (a\!+\!b,d) \overset{\color{#90f}{\rm E}}= (a\!+\!b,\,\color{#0a0}{d\bmod a\!+\!b})\,$ is as claimed, by $\!\bmod{\,\color{#0a0}{\!a\!+\!b}}\!:\ \underbrace{a^{\large k}\!+\!b^k \equiv b^k(\color{#0a0}{(-1)^{\large k}\!+\!1})}_{\large\ \ \ \color{#0a0}{ a\ \,\equiv\ -b}\ \ \ \ \ }^{\phantom .}\,$ and $\,(d,b^k\color{#0a0}e)=(d,\color{#0a0}e)\,$ by $\,(d,b)=1$.

Remark $ $ We can easily extend the above to the case when $\,m,n\,$ are not coprime.

Corollary $\ $ If $\,(A,B)=1\,$ and $\,M,N\in \Bbb N,\,$ and wlog $\,M/(M,N)\,$ is odd, then

$\quad(A^M\!+\!B^M,A^N\!+\!B^N)\, =\, (A^{(M,N)}\!+\!B^{(M,N)},C),\,\ \begin{cases} C = 2\ \ {\rm if}\ \ 2\mid N/(M,N)\\ C = 0\ \ {\rm otherwise}\end{cases}$

Proof $\ $ Let $\,D = (M,N),\,\ a = A^D,\ b = B^D.\,$ Then $\,(m,n) := (M/D,\:\! N/D) = 1\,$ and

$\quad\begin{align} (A^{M}\!+\!B^{M},A^{N}\!+\!B^{N})\, &=\, {(A^{D\large m}\!+\!B^{D\large m},(A^{D\large n}\!+\!B^{D\large n})}\\[.2em] &=\,{ \ \ \ (a^{\large m}\ + \ b^{\large m},\, \ \ \ \ a^{\large n}\ + \ b^{\large n})}\end{align}\ $ so the Theorem applies.

Nice solution, although was kind of difficult to read. Especially the part $(d,b)=(a^m,a^n,b)=1$ [I can prove it's true but doesn't seem obvious]. Also maybe you could add one more line to finish. I would write it like this: we got $d|a+b$, $d=(a^m+b^m,a^n+b^n)=(a+b,a^n+b^n)$ since $a+b|a^m+b^m$ [here we also use $m$ is odd]. And then we replace $a^n+b^n$ mod $a+b$. — Julian Mejia, May 24 '20 at 00:07
@Julian It is simply Euclidean gcd reduction $,(a_1,a_2,b) = (\bar a_1,\bar a_2,,b)\ $ if $\ \color{#0a0}{a_i\equiv \bar a_i} \pmod{!b}.$ Thus $$,(d,b) =({a^m!+b^m},{a^n!+b^n},b) = (a^m,a^n,b) = 1\ \ {\rm by}\ \ (a,b)=1,\ \color{#0a0}{a^k+b^k\equiv a^k}!!!!\pmod{!b}$$ and similarly in the final line. Once we master basic laws like this we can write proofs more concisely as above - so the global structure of the argument is not obfuscated by nitty-gritty details... — Bill Dubuque, May 24 '20 at 00:53
... (just like we use basic arithmetic laws (associative, commutative, distributive) without explicit mention in ordinary (ring) arithmetic). — Bill Dubuque, May 24 '20 at 00:58

score -1 · Answer 3 · answered May 23 '20 at 06:25

-1

When $m$ and $n$ are both odd and coprime and $a,b$ are coprime, $gcd(a^m + b^m, a^n + b^n) = a + b$.

As stated by @Geoffrey in the comments, $a+b \mid a^k + b^k$ for odd $k$:

$$a^m+b^m = (a + b)\underbrace{(a^{m-1} - a^{m-2}b + a^{m-3}b^{2} - \dots + b^{m-1})}_{m\space terms}$$ $$a^n+b^n = (a + b)\underbrace{(a^{n-1} - a^{n-2}b + a^{n-3}b^{2} - \dots + b^{n-1})}_{n\space terms}$$

The rightmost polynomials in both equations have no common factor since $m$ and $n$ have no common factor.

If $m,n$ are both even, then they cannot be coprime. If one of $m, n$ is even, then...?

answered May 23 '20 at 06:25

David Diaz

2,218

1

But you have not proved any of your claims. – Bill Dubuque May 23 '20 at 20:45
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@Gone Havent I? There arent many claims here, but how do you propose I support them? Also, what is a voting irregularity? And how come youve never asked a question? Do you have another account for that? Is that what led to your voting irregularities? Honest questions, please dont read malice into them – David Diaz May 25 '20 at 23:00
I got the claims you mentioned as well after experimentation with the exponents of $5$ and $7$... but I got stumped by the problem because don't know how to prove these claims. You have to prove them in order for the solution to be rigorous and correct. – Aarony Jamesys Jul 30 '20 at 04:33

Express $ \operatorname{gcd}\left(5^{m}+7^{m}, 5^{n}+7^{n}\right) $ in terms of $m$ and $n$

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