TITLE: $0 \leq a^2 + b^2 - abc \leq c \implies a^2 + b^2 - abc$ is a perfect square
QUESTION [6 upvotes]: 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.

REPLY [4 votes]: 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.