Can the sum, difference and product of 2 numbers be perfect squares?
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?
mathematics number-theory algebra
New contributor
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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?
mathematics number-theory algebra
New contributor
Is this on topic? Feels like a math question
– Dr Xorile
1 hour ago
No I made it myself but could not solve it.
– Magic turtle
1 hour ago
add a comment |
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?
mathematics number-theory algebra
New contributor
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?
mathematics number-theory algebra
mathematics number-theory algebra
New contributor
New contributor
edited 2 hours ago
Magic turtle
New contributor
asked 3 hours ago
Magic turtleMagic turtle
205
205
New contributor
New contributor
Is this on topic? Feels like a math question
– Dr Xorile
1 hour ago
No I made it myself but could not solve it.
– Magic turtle
1 hour ago
add a comment |
Is this on topic? Feels like a math question
– Dr Xorile
1 hour ago
No I made it myself but could not solve it.
– Magic turtle
1 hour ago
Is this on topic? Feels like a math question
– Dr Xorile
1 hour ago
Is this on topic? Feels like a math question
– Dr Xorile
1 hour ago
No I made it myself but could not solve it.
– Magic turtle
1 hour ago
No I made it myself but could not solve it.
– Magic turtle
1 hour ago
add a comment |
2 Answers
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The answer is
that they cannot.
Suppose $x+y=a^2$ and $x-y=b^2$. Then
$x,y=frac{a^2pm b^2}2$ 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.
add a comment |
(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
2 hours ago
But can x be 0?
– Ian MacDonald
1 hour ago
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
1 hour ago
add a comment |
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2 Answers
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2 Answers
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The answer is
that they cannot.
Suppose $x+y=a^2$ and $x-y=b^2$. Then
$x,y=frac{a^2pm b^2}2$ 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.
add a comment |
The answer is
that they cannot.
Suppose $x+y=a^2$ and $x-y=b^2$. Then
$x,y=frac{a^2pm b^2}2$ 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.
add a comment |
The answer is
that they cannot.
Suppose $x+y=a^2$ and $x-y=b^2$. Then
$x,y=frac{a^2pm b^2}2$ 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.
The answer is
that they cannot.
Suppose $x+y=a^2$ and $x-y=b^2$. Then
$x,y=frac{a^2pm b^2}2$ 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.
answered 1 hour ago
Gareth McCaughan♦Gareth McCaughan
60.6k3151235
60.6k3151235
add a comment |
add a comment |
(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
2 hours ago
But can x be 0?
– Ian MacDonald
1 hour ago
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
1 hour ago
add a comment |
(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
2 hours ago
But can x be 0?
– Ian MacDonald
1 hour ago
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
1 hour ago
add a comment |
(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)
(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)
edited 1 hour ago
answered 3 hours ago
Excited RaichuExcited Raichu
6,11821065
6,11821065
Sorry my fault, I meant to include that y cannot = 0.
– Magic turtle
2 hours ago
But can x be 0?
– Ian MacDonald
1 hour ago
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
1 hour ago
add a comment |
Sorry my fault, I meant to include that y cannot = 0.
– Magic turtle
2 hours ago
But can x be 0?
– Ian MacDonald
1 hour ago
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
1 hour ago
Sorry my fault, I meant to include that y cannot = 0.
– Magic turtle
2 hours ago
Sorry my fault, I meant to include that y cannot = 0.
– Magic turtle
2 hours ago
But can x be 0?
– Ian MacDonald
1 hour ago
But can x be 0?
– Ian MacDonald
1 hour ago
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
1 hour ago
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
1 hour ago
add a comment |
Magic turtle is a new contributor. Be nice, and check out our Code of Conduct.
Magic turtle is a new contributor. Be nice, and check out our Code of Conduct.
Magic turtle is a new contributor. Be nice, and check out our Code of Conduct.
Magic turtle is a new contributor. Be nice, and check out our Code of Conduct.
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Is this on topic? Feels like a math question
– Dr Xorile
1 hour ago
No I made it myself but could not solve it.
– Magic turtle
1 hour ago