Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The question reads as follows:

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?

share|cite|improve this question
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
up vote 1 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.