Can the sum, difference and product of 2 numbers be perfect squares? [on hold]
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 yesterday
Magic turtleMagic turtle
566
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put on hold as off-topic by rhsquared, Glorfindel, JonMark Perry, Excited Raichu, ABcDexter yesterday
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 |
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 yesterday
Magic turtleMagic turtle
566
New contributor
Magic turtle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by rhsquared, Glorfindel, JonMark Perry, Excited Raichu, ABcDexter yesterday
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
Is this on topic? Feels like a math question
– Dr Xorile
yesterday
No I made it myself but could not solve it.
– Magic turtle
yesterday
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
asked yesterday
Magic turtleMagic turtle
566
New contributor
Magic turtle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 yesterday
Magic turtleMagic turtle
566
New contributor
Magic turtle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Magic turtleMagic turtle
566
New contributor
Magic turtle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Magic turtle
asked yesterday
Magic turtleMagic turtle
566
New contributor
Magic turtle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Magic turtleMagic turtle
566
asked yesterday
Magic turtleMagic turtle
566
566
New contributor
Magic turtle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Magic turtle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Magic turtle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by rhsquared, Glorfindel, JonMark Perry, Excited Raichu, ABcDexter yesterday
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.
put on hold as off-topic by rhsquared, Glorfindel, JonMark Perry, Excited Raichu, ABcDexter yesterday
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
Is this on topic? Feels like a math question
– Dr Xorile
yesterday
No I made it myself but could not solve it.
– Magic turtle
yesterday
add a comment |
1
Is this on topic? Feels like a math question
– Dr Xorile
yesterday
No I made it myself but could not solve it.
– Magic turtle
yesterday
1
1
Is this on topic? Feels like a math question
– Dr Xorile
yesterday
Is this on topic? Feels like a math question
– Dr Xorile
yesterday
No I made it myself but could not solve it.
– Magic turtle
yesterday
No I made it myself but could not solve it.
– Magic turtle
yesterday
add a comment |
2 Answers
2
active
oldest
votes
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 yesterday
Gareth McCaughan♦Gareth McCaughan
60.7k3152235
"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
yesterday
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♦
yesterday
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)
answered yesterday
Excited RaichuExcited Raichu
6,11821065
Sorry my fault, I meant to include that y cannot = 0.
– Magic turtle
yesterday
But can x be 0?
– Ian MacDonald
yesterday
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
yesterday
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 yesterday
Gareth McCaughan♦Gareth McCaughan
60.7k3152235
"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
yesterday
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♦
yesterday
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.
answered yesterday
Gareth McCaughan♦Gareth McCaughan
60.7k3152235
"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
yesterday
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♦
yesterday
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.
answered yesterday
Gareth McCaughan♦Gareth McCaughan
60.7k3152235
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 yesterday
Gareth McCaughan♦Gareth McCaughan
60.7k3152235
answered yesterday
Gareth McCaughan♦Gareth McCaughan
60.7k3152235
answered yesterday
Gareth McCaughan♦Gareth McCaughan
60.7k3152235
answered yesterday
Gareth McCaughan♦Gareth McCaughan
60.7k3152235
60.7k3152235
"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
yesterday
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♦
yesterday
add a comment |
"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
yesterday
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♦
yesterday
"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
yesterday
"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
yesterday
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♦
yesterday
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♦
yesterday
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)
answered yesterday
Excited RaichuExcited Raichu
6,11821065
Sorry my fault, I meant to include that y cannot = 0.
– Magic turtle
yesterday
But can x be 0?
– Ian MacDonald
yesterday
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
yesterday
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)
answered yesterday
Excited RaichuExcited Raichu
6,11821065
Sorry my fault, I meant to include that y cannot = 0.
– Magic turtle
yesterday
But can x be 0?
– Ian MacDonald
yesterday
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
yesterday
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)
answered yesterday
Excited RaichuExcited Raichu
6,11821065
(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 yesterday
Excited RaichuExcited Raichu
6,11821065
edited yesterday
answered yesterday
Excited RaichuExcited Raichu
6,11821065
answered yesterday
Excited RaichuExcited Raichu
6,11821065
answered yesterday
Excited RaichuExcited Raichu
6,11821065
6,11821065
Sorry my fault, I meant to include that y cannot = 0.
– Magic turtle
yesterday
But can x be 0?
– Ian MacDonald
yesterday
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
yesterday
add a comment |
Sorry my fault, I meant to include that y cannot = 0.
– Magic turtle
yesterday
But can x be 0?
– Ian MacDonald
yesterday
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
yesterday
Sorry my fault, I meant to include that y cannot = 0.
– Magic turtle
yesterday
Sorry my fault, I meant to include that y cannot = 0.
– Magic turtle
yesterday
But can x be 0?
– Ian MacDonald
yesterday
But can x be 0?
– Ian MacDonald
yesterday
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
yesterday
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
yesterday
add a comment |
1
Is this on topic? Feels like a math question
– Dr Xorile
yesterday
No I made it myself but could not solve it.
– Magic turtle
yesterday