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 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.

share|cite|improve this question
The inequalities imply that $c$ must lie between $(a^2+b^2)/(1+ab)$ and $(a^2+b^2)/ab$. For any $a, b \ge 1$, there is at most one $c$ that can satisfy those conditions. I'm not sure if that leads anywhere. – Rahul Jun 17 '12 at 6:43
Interesting problem. I have verified it experimentally for a,b,c up to 100. – imallett Jun 17 '12 at 6:50
I edited your title to reflect the content of the question. – mixedmath Jun 17 '12 at 6:58
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. – Rahul Jun 17 '12 at 8:11

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.

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.