If we take 2 numbers $x$ and $y$ such that $x>y>0$ and , can $x + y$, $x - y$ and $xy$ all be perfect squares?


closed as off-topic by rhsquared, Glorfindel, JonMark Perry, Excited Raichu, ABcDexter Jan 7 at 16:54

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  • 1
    $\begingroup$ Is this on topic? Feels like a math question $\endgroup$ – Dr Xorile Jan 7 at 15:05
  • $\begingroup$ No I made it myself but could not solve it. $\endgroup$ – Magic turtle Jan 7 at 15:06

The answer is

that they cannot.

Suppose $x+y=a^2$ and $x-y=b^2$. Then

$x,y=\frac{a^2\pm b^2}2$ and therefore $4xy=a^4-b^4$. So if this too is a square we have $a^4-b^4=c^2$. Now see https://math.stackexchange.com/questions/153546/solving-x4-y4-z2 where it is shown that there are no solutions to this equation in nonzero integers.

  • $\begingroup$ "if this too is a square" we have (a^4 -b^4) / 4 = c^2 ... The fact that you binned y=0 means that the proof doesn't hold for this alternate equation $\endgroup$ – UKMonkey Jan 7 at 17:10
  • $\begingroup$ If 4xy is a square then xy is also a square. I'm not sure what you mean by "you binned y=0" but the question itself specifies that y>0. $\endgroup$ – Gareth McCaughan Jan 7 at 18:56

(Before the OP clarifying that 0 is disallowed)


x = 4, y = 0; x+y = 4, x-y = 4, xy = 0. (x can be any perfect square, I just decided to use 4)

  • $\begingroup$ Sorry my fault, I meant to include that y cannot = 0. $\endgroup$ – Magic turtle Jan 7 at 14:10
  • $\begingroup$ But can x be 0? $\endgroup$ – Ian MacDonald Jan 7 at 14:17
  • $\begingroup$ If x were 0 y would be less than 0 making x + y negative and a negative number cannot be a perfect square $\endgroup$ – Magic turtle Jan 7 at 14:30

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