Let $x, y$ and $z$ be positive integers with $x<y$. It appears that the integer $$(y^2z^3-x)(y^2z^3+3x)$$ is never a perfect square. Why? A proof? I'm not sure if it is easy.
$\begingroup$
$\endgroup$
4
asked Sep 18, 2016 at 12:49
-
$\begingroup$ Note: a "perfect square" and "perfect number" are two very different notions. You mean the former one here. $\endgroup$– WojowuCommented Sep 18, 2016 at 12:54
-
$\begingroup$ True. Corrected. $\endgroup$– T. AmdeberhanCommented Sep 18, 2016 at 12:59
-
3$\begingroup$ Looks like a hometask. Search for a square close to your number... $\endgroup$– Ilya BogdanovCommented Sep 18, 2016 at 13:20
-
$\begingroup$ @IlyaBogdanov This question is a corrected version of one previously asked by the OP, where he said that the original motivation was from some calculation in a real analysis problem $\endgroup$– Yemon ChoiCommented Sep 18, 2016 at 14:49
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
1
Rewrite your expression as $(y^2z^3+x)^2-4x^2$. This is clearly less than $(y^2z^3+x)^2$. On the other hand, it is larger than $(y^2z^3+x-2)^2=(y^2z^3+x)^2-4(y^2z^3+x)+4$, since $4(y^2z^3+x)-4\geq 4y^2+4x-4\geq 4y^2>4x^2$. So if this expression is a square, it must be $(y^2z^3+x-1)^2=(y^2z^3+x)^2-2(y^2z^3+x)+1$, so $4x^2=2(y^2z^3+x)-1$. LHS is even, while RHS is odd, so this can't be.
-
$\begingroup$ Cute. Thanks. Actually it was easy then. $\endgroup$ Commented Sep 18, 2016 at 13:54