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Source: Romania TST 1993

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KDS

#1 h Jul 12, 2009, 10:45 AM • 1 Y

Y by Adventure10

Prove that the equation $(x+y)^n=x^m+y^m$ has a unique solution in integers with $x>y>0$ and $m,n>1$ .

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shoki

#2 h Aug 29, 2009, 11:57 PM • 2 Y

Y by Adventure10, Mango247

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#3 h Apr 23, 2022, 6:08 PM • 3 Y

Y by Mango247, Mango247, Mango247

I will instead characterize every solution $(x,y,m,n)$ given that $x,y>0$ and $m,n \geq 1$ —this is the version of the problem I received.

Clearly $(x,y,m,n)=(x,y,1,1)$ works, and by size we require $n\geq m$ hence no other solution exists for $n=1$ , so assume $n>1$ . Then as a consequence of Zsigmondy there must exist some prime dividing $x^n+y^n$ that doesn't divide $x+y$ and thus $(x+y)^m$ , unless:

$(x,y)=(k,2k)$ for some $k$ , and $n=3$ hence $m \in \{1,2\}$ . Then either $3k=9k^3$ which is impossible or $9k^2=9k^3 \implies k=1$ , which yields $(x,y,m,n)=(1,2,2,3)$ and its permutation $(2,1,2,3)$ .
$(x,y)=(k,k)$ for some $k$ , so $2^mk^m=2k^n$ from which we find that $k$ must be a power of $2$ , say $2^j$ . Then
$2^{m+mj}=2^{nj+1} \implies j=\frac{m-1}{n-m},$ from which we extract the very ugly answer $\left(2^\frac{m-1}{n-m},2^\frac{m-1}{n-m},m,n\right)$ for all $(m,n)$ such that $n-m \mid m-1$ .

Since everything is reversible, it follows that the solutions are $(x,y,m,n)=(x,y,1,1)$ $,(1,2,2,3),(2,1,2,3)$ , and $\left(2^\frac{m-1}{n-m},2^\frac{m-1}{n-m},m,n\right)$ for all $(m,n)$ such that $n-m \mid m-1$ . $\blacksquare$ (clearly, this implies the original problem)

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#4 h Nov 15, 2023, 12:43 AM

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probably wrong sol

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#5 h Apr 10, 2025, 3:22 AM

Y by

Nearly identical to #3, so this is for storage.

Quote:

Solve in positive integers the equation $(x+y)^m=x^n+y^n.$

Observe that $m=1 \iff n=1,$ and this clearly yields a solution. So assume that $m, n \geq 2.$ Observe that by size $n > m.$

Now, if $x=y=k$ we get $2^nk^n=2k^m \iff k=2^{\frac{m-1}{n-m}},$ which indeed is a solution as long as $(n-m) | (m-1).$

WLOG assume that $x>y.$ Then let $d=\gcd(x, y), x=da, y=db.$ We get $d^m(a+b)^m = d^n(a^n+b^n) \implies (a+b)^m = d^{n-m} (a^n+b^n).$ By Zsigmondy's Theorem there is prime dividing $a^n+b^n$ but not $a+b,$ which is a contradiction, unless we have $2^3+1$ in which $a=2, b=1, n=3.$ This would give $3^m=d^{3-m} \cdot 9,$ and testing possibilities yields $m=2$ so $(x, y, m, n) = (1, 2, 2, 3), (2, 1, 2, 3).$

To summarize, the solutions are $(x, y, m, n) = (x, y, 1, 1), (1, 2, 2, 3), (2, 1, 2, 3), \left( 2^{\frac{m-1}{n-m}}, 2^{\frac{m-1}{n-m}}, m, n \right)$ for $(m, n)$ that yields integer values for the last one.

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