6
$\begingroup$

The following problem looks very interesting to me and I cannot even guess a solution to it. It states that:

Suppose that $a,b,c$ are three natural numbers satisfying the inequality: $0\leq a^2 + b^2 - abc\leq c$. Show that $a^2 + b^2 - abc$ is a perfect square.

Cases like $a=b$ or $a=1$ can be handled very easily, but is there any general solution? Any help shall be greatly appreciated.

  • 0
    Regarding my previous comment, make that *at most two* values of $c$. I had checked that the length of the range was at most $1$ and jumped to a conclusion, but you can have $a=1$, $b=1$, $c\in\{1,2\}$. I still think for $(a,b)\ne(1,1)$ there can't be more than one solution, though.2012-06-17

1 Answers 1

4

A solution appears in the first link in this question:

Seemingly invalid step in the proof of $\frac{a^2+b^2}{ab+1}$ is a perfect square?

This is a variant of the $ab+1$ problem, maybe devised by working backward from the solution.