# Can the sum, difference and product of 2 numbers be perfect squares? [closed]

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, ABcDexterJan 7 at 16:54

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

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.

• "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 – UKMonkey Jan 7 at 17:10
• 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. – Gareth McCaughan Jan 7 at 18:56

(Before the OP clarifying that 0 is disallowed)

Yes.

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

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