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Perfect square of rational number
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This question is asked in the mathematics contest for grade 9.
Prove that if $x,y$ are two rational numbers such that $x^3+y^3=2xy$ then $1-xy$ is a perfect square of a rational number.
number-theory elementary-number-theory
asked 57 mins ago
Blind
503217
add a comment |Â
up vote
5
down vote
favorite
This question is asked in the mathematics contest for grade 9.
Prove that if $x,y$ are two rational numbers such that $x^3+y^3=2xy$ then $1-xy$ is a perfect square of a rational number.
number-theory elementary-number-theory
asked 57 mins ago
Blind
503217
2
Maybe you can start by replacing $x$ by $a/b$ and $y$ by $p/q$ and see what equations can come out
– Mark
49 mins ago
1
Note that $x=1$ and $y=1$ is a solution! I suspect this is the only rational solution, in which case $1-xy=0^2$ does in fact hold. The hard part is proving that it is the only rational solution.
– Zubin Mukerjee
35 mins ago
I have an idea to substitute 1 by $fracx^3+y^32xy $ into the expression $1-xy$ to get a perfect square of a rational number but I cannot go on.
– Blind
35 mins ago
How about x=y=0?
– Blind
33 mins ago
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
This question is asked in the mathematics contest for grade 9.
Prove that if $x,y$ are two rational numbers such that $x^3+y^3=2xy$ then $1-xy$ is a perfect square of a rational number.
number-theory elementary-number-theory
asked 57 mins ago
Blind
503217
This question is asked in the mathematics contest for grade 9.
Prove that if $x,y$ are two rational numbers such that $x^3+y^3=2xy$ then $1-xy$ is a perfect square of a rational number.
number-theory elementary-number-theory
number-theory elementary-number-theory
asked 57 mins ago
Blind
503217
asked 57 mins ago
Blind
503217
edited 35 mins ago
asked 57 mins ago
Blind
503217
asked 57 mins ago
Blind
503217
asked 57 mins ago
Blind
503217
503217
2
Maybe you can start by replacing $x$ by $a/b$ and $y$ by $p/q$ and see what equations can come out
– Mark
49 mins ago
1
Note that $x=1$ and $y=1$ is a solution! I suspect this is the only rational solution, in which case $1-xy=0^2$ does in fact hold. The hard part is proving that it is the only rational solution.
– Zubin Mukerjee
35 mins ago
I have an idea to substitute 1 by $fracx^3+y^32xy $ into the expression $1-xy$ to get a perfect square of a rational number but I cannot go on.
– Blind
35 mins ago
How about x=y=0?
– Blind
33 mins ago
add a comment |Â
2
Maybe you can start by replacing $x$ by $a/b$ and $y$ by $p/q$ and see what equations can come out
– Mark
49 mins ago
1
Note that $x=1$ and $y=1$ is a solution! I suspect this is the only rational solution, in which case $1-xy=0^2$ does in fact hold. The hard part is proving that it is the only rational solution.
– Zubin Mukerjee
35 mins ago
I have an idea to substitute 1 by $fracx^3+y^32xy $ into the expression $1-xy$ to get a perfect square of a rational number but I cannot go on.
– Blind
35 mins ago
How about x=y=0?
– Blind
33 mins ago
2
2
Maybe you can start by replacing $x$ by $a/b$ and $y$ by $p/q$ and see what equations can come out
– Mark
49 mins ago
Maybe you can start by replacing $x$ by $a/b$ and $y$ by $p/q$ and see what equations can come out
– Mark
49 mins ago
1
1
Note that $x=1$ and $y=1$ is a solution! I suspect this is the only rational solution, in which case $1-xy=0^2$ does in fact hold. The hard part is proving that it is the only rational solution.
– Zubin Mukerjee
35 mins ago
Note that $x=1$ and $y=1$ is a solution! I suspect this is the only rational solution, in which case $1-xy=0^2$ does in fact hold. The hard part is proving that it is the only rational solution.
– Zubin Mukerjee
35 mins ago
I have an idea to substitute 1 by $fracx^3+y^32xy $ into the expression $1-xy$ to get a perfect square of a rational number but I cannot go on.
– Blind
35 mins ago
I have an idea to substitute 1 by $fracx^3+y^32xy $ into the expression $1-xy$ to get a perfect square of a rational number but I cannot go on.
– Blind
35 mins ago
How about x=y=0?
– Blind
33 mins ago
How about x=y=0?
– Blind
33 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
5
down vote
accepted
The result is trivial unless $x,,y$ are non-zero, so that $y=tx$ with $t$ a non-zero rational number. Then $x^3(1+t^3)=2tx^2$ and $$x=frac2t1+t^3,,1-xy=frac(1+t^3)^2-4t^3(1+t^3)^2=(frac1-t^31+t^3)^2.$$The division only requires $1+t^3ne 0$, which in turn trivially follows from $x^3+y^3=2xyne 0$.
answered 30 mins ago
J.G.
14.9k11626
Nice answer. Thanks
– Blind
24 mins ago
Based on the proof of J.G., I propose another solution
– Blind
13 mins ago
add a comment |Â
up vote
1
down vote
My solution is based on the idea of J.G.
$1-xy=1-fracx^3yx^2=fracx^2-x^3yx^2=fracx^2+y^4-2xy^2x^2=frac(x-y^2)^2x^2$.
I think that this is the shortest solution.
answered 9 mins ago
Blind
503217
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
The result is trivial unless $x,,y$ are non-zero, so that $y=tx$ with $t$ a non-zero rational number. Then $x^3(1+t^3)=2tx^2$ and $$x=frac2t1+t^3,,1-xy=frac(1+t^3)^2-4t^3(1+t^3)^2=(frac1-t^31+t^3)^2.$$The division only requires $1+t^3ne 0$, which in turn trivially follows from $x^3+y^3=2xyne 0$.
answered 30 mins ago
J.G.
14.9k11626
Nice answer. Thanks
– Blind
24 mins ago
Based on the proof of J.G., I propose another solution
– Blind
13 mins ago
add a comment |Â
up vote
5
down vote
accepted
The result is trivial unless $x,,y$ are non-zero, so that $y=tx$ with $t$ a non-zero rational number. Then $x^3(1+t^3)=2tx^2$ and $$x=frac2t1+t^3,,1-xy=frac(1+t^3)^2-4t^3(1+t^3)^2=(frac1-t^31+t^3)^2.$$The division only requires $1+t^3ne 0$, which in turn trivially follows from $x^3+y^3=2xyne 0$.
answered 30 mins ago
J.G.
14.9k11626
Nice answer. Thanks
– Blind
24 mins ago
Based on the proof of J.G., I propose another solution
– Blind
13 mins ago
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
The result is trivial unless $x,,y$ are non-zero, so that $y=tx$ with $t$ a non-zero rational number. Then $x^3(1+t^3)=2tx^2$ and $$x=frac2t1+t^3,,1-xy=frac(1+t^3)^2-4t^3(1+t^3)^2=(frac1-t^31+t^3)^2.$$The division only requires $1+t^3ne 0$, which in turn trivially follows from $x^3+y^3=2xyne 0$.
answered 30 mins ago
J.G.
14.9k11626
The result is trivial unless $x,,y$ are non-zero, so that $y=tx$ with $t$ a non-zero rational number. Then $x^3(1+t^3)=2tx^2$ and $$x=frac2t1+t^3,,1-xy=frac(1+t^3)^2-4t^3(1+t^3)^2=(frac1-t^31+t^3)^2.$$The division only requires $1+t^3ne 0$, which in turn trivially follows from $x^3+y^3=2xyne 0$.
answered 30 mins ago
J.G.
14.9k11626
answered 30 mins ago
J.G.
14.9k11626
answered 30 mins ago
J.G.
14.9k11626
answered 30 mins ago
J.G.
14.9k11626
14.9k11626
Nice answer. Thanks
– Blind
24 mins ago
Based on the proof of J.G., I propose another solution
– Blind
13 mins ago
add a comment |Â
Nice answer. Thanks
– Blind
24 mins ago
Based on the proof of J.G., I propose another solution
– Blind
13 mins ago
Nice answer. Thanks
– Blind
24 mins ago
Nice answer. Thanks
– Blind
24 mins ago
Based on the proof of J.G., I propose another solution
– Blind
13 mins ago
Based on the proof of J.G., I propose another solution
– Blind
13 mins ago
add a comment |Â
up vote
1
down vote
My solution is based on the idea of J.G.
$1-xy=1-fracx^3yx^2=fracx^2-x^3yx^2=fracx^2+y^4-2xy^2x^2=frac(x-y^2)^2x^2$.
I think that this is the shortest solution.
answered 9 mins ago
Blind
503217
add a comment |Â
up vote
1
down vote
My solution is based on the idea of J.G.
$1-xy=1-fracx^3yx^2=fracx^2-x^3yx^2=fracx^2+y^4-2xy^2x^2=frac(x-y^2)^2x^2$.
I think that this is the shortest solution.
answered 9 mins ago
Blind
503217
add a comment |Â
up vote
1
down vote
up vote
1
down vote
My solution is based on the idea of J.G.
$1-xy=1-fracx^3yx^2=fracx^2-x^3yx^2=fracx^2+y^4-2xy^2x^2=frac(x-y^2)^2x^2$.
I think that this is the shortest solution.
answered 9 mins ago
Blind
503217
My solution is based on the idea of J.G.
$1-xy=1-fracx^3yx^2=fracx^2-x^3yx^2=fracx^2+y^4-2xy^2x^2=frac(x-y^2)^2x^2$.
I think that this is the shortest solution.
answered 9 mins ago
Blind
503217
edited 2 mins ago
answered 9 mins ago
Blind
503217
answered 9 mins ago
Blind
503217
answered 9 mins ago
Blind
503217
503217
add a comment |Â
add a comment |Â
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2
Maybe you can start by replacing $x$ by $a/b$ and $y$ by $p/q$ and see what equations can come out
– Mark
49 mins ago
1
Note that $x=1$ and $y=1$ is a solution! I suspect this is the only rational solution, in which case $1-xy=0^2$ does in fact hold. The hard part is proving that it is the only rational solution.
– Zubin Mukerjee
35 mins ago
I have an idea to substitute 1 by $fracx^3+y^32xy $ into the expression $1-xy$ to get a perfect square of a rational number but I cannot go on.
– Blind
35 mins ago
How about x=y=0?
– Blind
33 mins ago