Mixing
Can the sum, difference and product of 2 numbers be perfect squares? [closed]
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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
asked Jan 7 at 12:55
Magic turtleMagic turtle
566
$endgroup$
closed as off-topic by rhsquared, Glorfindel, JonMark Perry, Excited Raichu, ABcDexter Jan 7 at 16:54
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – rhsquared, Glorfindel, JonMark Perry, Excited Raichu, ABcDexter
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
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
asked Jan 7 at 12:55
Magic turtleMagic turtle
566
$endgroup$
closed as off-topic by rhsquared, Glorfindel, JonMark Perry, Excited Raichu, ABcDexter Jan 7 at 16:54
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – rhsquared, Glorfindel, JonMark Perry, Excited Raichu, ABcDexter
If this question can be reworded to fit the rules in the help center, please edit the question.
1
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Is this on topic? Feels like a math question
$endgroup$
– Dr Xorile
Jan 7 at 15:05
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No I made it myself but could not solve it.
$endgroup$
– Magic turtle
Jan 7 at 15:06
add a comment |
$begingroup$
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
asked Jan 7 at 12:55
Magic turtleMagic turtle
566
$endgroup$
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
asked Jan 7 at 12:55
Magic turtleMagic turtle
566
asked Jan 7 at 12:55
Magic turtleMagic turtle
566
edited Jan 7 at 14:09
Magic turtle
asked Jan 7 at 12:55
Magic turtleMagic turtle
566
asked Jan 7 at 12:55
Magic turtleMagic turtle
566
asked Jan 7 at 12:55
Magic turtleMagic turtle
566
566
closed as off-topic by rhsquared, Glorfindel, JonMark Perry, Excited Raichu, ABcDexter Jan 7 at 16:54
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – rhsquared, Glorfindel, JonMark Perry, Excited Raichu, ABcDexter
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by rhsquared, Glorfindel, JonMark Perry, Excited Raichu, ABcDexter Jan 7 at 16:54
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – rhsquared, Glorfindel, JonMark Perry, Excited Raichu, ABcDexter
If this question can be reworded to fit the rules in the help center, please edit the question.
1
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Is this on topic? Feels like a math question
$endgroup$
– Dr Xorile
Jan 7 at 15:05
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No I made it myself but could not solve it.
$endgroup$
– Magic turtle
Jan 7 at 15:06
add a comment |
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
1
1
$begingroup$
Is this on topic? Feels like a math question
$endgroup$
– Dr Xorile
Jan 7 at 15:05
$begingroup$
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
$begingroup$
No I made it myself but could not solve it.
$endgroup$
– Magic turtle
Jan 7 at 15:06
add a comment |
2 Answers
2
active
oldest
votes
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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.
answered Jan 7 at 14:15
Gareth McCaughan♦Gareth McCaughan
61.1k3152236
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"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
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– UKMonkey
Jan 7 at 17:10
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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.
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– Gareth McCaughan♦
Jan 7 at 18:56
add a comment |
$begingroup$
(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)
answered Jan 7 at 12:56
Excited RaichuExcited Raichu
6,43521166
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Sorry my fault, I meant to include that y cannot = 0.
$endgroup$
– Magic turtle
Jan 7 at 14:10
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But can x be 0?
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– Ian MacDonald
Jan 7 at 14:17
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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
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 Jan 7 at 14:15
Gareth McCaughan♦Gareth McCaughan
61.1k3152236
$endgroup$
$begingroup$
"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
add a comment |
$begingroup$
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 Jan 7 at 14:15
Gareth McCaughan♦Gareth McCaughan
61.1k3152236
$endgroup$
$begingroup$
"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
add a comment |
$begingroup$
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 Jan 7 at 14:15
Gareth McCaughan♦Gareth McCaughan
61.1k3152236
$endgroup$
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 Jan 7 at 14:15
Gareth McCaughan♦Gareth McCaughan
61.1k3152236
answered Jan 7 at 14:15
Gareth McCaughan♦Gareth McCaughan
61.1k3152236
answered Jan 7 at 14:15
Gareth McCaughan♦Gareth McCaughan
61.1k3152236
answered Jan 7 at 14:15
Gareth McCaughan♦Gareth McCaughan
61.1k3152236
61.1k3152236
$begingroup$
"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
add a comment |
$begingroup$
"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
$begingroup$
"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 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
$begingroup$
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
add a comment |
$begingroup$
(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)
answered Jan 7 at 12:56
Excited RaichuExcited Raichu
6,43521166
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
(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)
answered Jan 7 at 12:56
Excited RaichuExcited Raichu
6,43521166
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
(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)
answered Jan 7 at 12:56
Excited RaichuExcited Raichu
6,43521166
$endgroup$
(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)
answered Jan 7 at 12:56
Excited RaichuExcited Raichu
6,43521166
edited Jan 7 at 14:17
answered Jan 7 at 12:56
Excited RaichuExcited Raichu
6,43521166
answered Jan 7 at 12:56
Excited RaichuExcited Raichu
6,43521166
answered Jan 7 at 12:56
Excited RaichuExcited Raichu
6,43521166
6,43521166
$begingroup$
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
add a comment |
$begingroup$
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
$begingroup$
Sorry my fault, I meant to include that y cannot = 0.
$endgroup$
– Magic turtle
Jan 7 at 14:10
$begingroup$
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
$begingroup$
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
$begingroup$
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
add a comment |
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