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Let $n$ be a positive integer. Show that the equation

$\displaystyle x^n + y^n = z^{n+1}$

has infinitely many integer solutions.

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2 Answers 2

Hint: It is pretty easy to find infinitely many with $x|y$.

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So, let y = x*m. Then x^n + (x*m)^n = x^n + (x^n)*(m^n) = (x^n)*((m^n) + 1) = z*z^n..... Is this the jist of it? THen you could say that if m^n + 1 = z for all m,n in Z and if x|y, then there are infinitely many solution? –  Eleven-Eleven Feb 20 '13 at 19:38
    
@ChristopherErnst So, now can you find a specific $x$ given $m$? It's sort of a generalization of the obviojus solution $x=y=z=2$ –  Thomas Andrews Feb 20 '13 at 19:39
    
So I'm thinking since we have this x^n(1+ m^n) = z^n *z, if z = 1 + m^n and x|y such that y = xm we can find an infinite amount of solutions since there are infinitely many m's to try? –  Eleven-Eleven Feb 20 '13 at 19:47
    
If you wanted $Ax^n$ to be a power of $n+1$ what is the most obvious value you could choose for $x$? –  Thomas Andrews Feb 20 '13 at 19:54
    
There are also some other infinite classes that are perhaps even more obvious - cases when $y=0$. –  Thomas Andrews Feb 20 '13 at 19:54
up vote 0 down vote accepted

Let $x$ and $y$ be such that $x \mid y$. This means that $y = x \cdot m$ where $m$ is a positive integer. Thus $x^n + y^n = x^n + (x \cdot m)^n = x^n + x^n \cdot m^n = x^n \cdot [1 + m^n] = z \cdot z^n = z^{n+1}$

So we know that when $x \mid y$, we have $x^n \cdot [1 + m^n] = z^n \cdot z$ Now let $x^n = z^n$. Thus $1 + m^n = z$ Then $z^n = [1 + m^n]^n = x^n$.

Substituting $[1 + m^n]^n$ for $x^n$ we get $$x^n + y^n = [1 + m^n]^n + (m[1 + m^n])^n = [1 + m^n]^n + m^n \cdot [1 + m^n]^n \\ = [1 + m^n] \cdot [1 + m^n]^n = [1 + m^n]^{n+1} = x^{n+1} = z^{n+1}$$

Since there are infinitely many choices for $m$ since $n$ is a positive integer, we have our answer; a triple that gives infinitely many solutions to the equation

$x^n + y^n = z^{n+1}$

is $(1 + m^n, m \cdot (1 + m^n), 1 + m^n)$.

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