# Number of solutions to equation.

Let $p$ and $k$ be positive integers such that $p$ is prime and $k > 1$. Prove that there is at most one pair $(x, y)$ of positive integers such that $x^k + px = y^k$.

I worked $x$ to be $1$ as $x = (y^k-x^k)/p$ where $p$ is prime. Therefore for $x$ to be real $y^k - x^k$ must be $p$. However I cannot prove this for $y$. Any suggestions?

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Parentheses, please. $x=y^k-x^k/p$ is incorrect. You should have $x=(y^k-x^k)/p$ You don't know that $y^k-x^k=p$, just that $p$ divides $y^k-x^k$ –  Ross Millikan Sep 25 '12 at 15:46
ah, I saw that now. –  fosho Sep 25 '12 at 15:50
$x$ doesn't have to be $1$.. take $x^2 + 5x = y^2$ which has solution $(4,6)$. –  Cocopuffs Sep 25 '12 at 17:17
Can you do it if you assume that x and p are relatively prime? –  deinst Sep 26 '12 at 17:39
can someone please show working? –  fosho Sep 27 '12 at 21:17

On the hypotheses, suppose $$x^k+px=y^k$$ Write this as $$x(x^{k-1}+p)=y^k$$ Anything that divides both $x$ and $x^{k-1}+p$ must divide $p$, hence, must be 1 or $p$.
First case: $\gcd(x,x^{k-1}+p)=1$. Then we have two relatively prime numebrs whose product is a $k$th power, so (by unique factorization) each must be a $k$th power; $$x=r^k,\quad x^{k-1}+p=s^k$$ for some $r$ and $s$. Putting these together, $$p=s^k-(r^{k-1})^k=(s-r^{k-1})(s^{k-1}+\cdots+r^{(k-1)^2})$$ from which we deduce $s=1+r^{k-1}$. This gives us an equation of the form $p=f(r)$ where $f$ is a polynomial with positive coefficients, hence an increasing function, hence, for any given $p$, there is at most one value of $r$ satisfying $f(r)=p$. Thus, there is at most one value of $x$, and at most one value of $y$.
Second case: $\gcd(x,x^{k-1}+p)=p$. Then either $$x=pr^k,\quad x^{k-1}+p=p^{k-1}s^k\tag1$$ or $$x=p^{k-1}r^k,\quad x^{k-1}+p=ps^k\tag2$$ From (1), we deduce $$p^{k-2}r^{k(k-1)}+1=p^{k-2}s^k$$ which implies $k=2$ and thus $r^2+1=s^2$, which is absurd.
From (2), we deduce $$s^k-(p^{k-2}r^{k-1})^k=1$$ which is also absurd, so we're done.