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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?

  • 0
    can someone please show working?2012-09-27

1 Answers 1

1

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.