Solution of Diophantine Equation $1+a^6=x^2$
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Does the equation $1+a^6=x^2$ have any other integer solutions except the trivial one $a=0,x=1$?
number-theory diophantine-equations
edited Dec 31 '18 at 7:56
dmtri
1,7652521
asked Dec 31 '18 at 6:32
ershersh
603113
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add a comment |
$begingroup$
Does the equation $1+a^6=x^2$ have any other integer solutions except the trivial one $a=0,x=1$?
number-theory diophantine-equations
edited Dec 31 '18 at 7:56
dmtri
1,7652521
asked Dec 31 '18 at 6:32
ershersh
603113
$endgroup$
add a comment |
$begingroup$
Does the equation $1+a^6=x^2$ have any other integer solutions except the trivial one $a=0,x=1$?
number-theory diophantine-equations
edited Dec 31 '18 at 7:56
dmtri
1,7652521
asked Dec 31 '18 at 6:32
ershersh
603113
$endgroup$
Does the equation $1+a^6=x^2$ have any other integer solutions except the trivial one $a=0,x=1$?
number-theory diophantine-equations
number-theory diophantine-equations
edited Dec 31 '18 at 7:56
dmtri
1,7652521
asked Dec 31 '18 at 6:32
ershersh
603113
edited Dec 31 '18 at 7:56
dmtri
1,7652521
asked Dec 31 '18 at 6:32
ershersh
603113
edited Dec 31 '18 at 7:56
dmtri
1,7652521
edited Dec 31 '18 at 7:56
dmtri
1,7652521
edited Dec 31 '18 at 7:56
dmtri
1,7652521
1,7652521
asked Dec 31 '18 at 6:32
ershersh
603113
asked Dec 31 '18 at 6:32
ershersh
603113
asked Dec 31 '18 at 6:32
ershersh
603113
603113
add a comment |
add a comment |
2 Answers
2
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oldest
votes
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It's $$(x-a^3)(x+a^3)=1$$ and solve two systems:
$$x-a^3=1$$ and $$x+a^3=1$$ or
$$x-a^3=-1$$ and $$x+a^3=-1,$$
which gives the answer for $(x,a)$:
$${(1,0), (-1,0)}.$$
answered Dec 31 '18 at 7:21
Michael RozenbergMichael Rozenberg
111k1897201
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$x^2$ and $a^6$ are both perfect squares. And the only two perfect squares that differ by $1$ are $0$ and $1$ (because for any $nge 0$, the next biggest square after $n^2$ is $n^2+2n+1$).
answered Dec 31 '18 at 7:33
TonyKTonyK
44.1k358137
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2 Answers
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2 Answers
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$begingroup$
It's $$(x-a^3)(x+a^3)=1$$ and solve two systems:
$$x-a^3=1$$ and $$x+a^3=1$$ or
$$x-a^3=-1$$ and $$x+a^3=-1,$$
which gives the answer for $(x,a)$:
$${(1,0), (-1,0)}.$$
answered Dec 31 '18 at 7:21
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
add a comment |
$begingroup$
It's $$(x-a^3)(x+a^3)=1$$ and solve two systems:
$$x-a^3=1$$ and $$x+a^3=1$$ or
$$x-a^3=-1$$ and $$x+a^3=-1,$$
which gives the answer for $(x,a)$:
$${(1,0), (-1,0)}.$$
answered Dec 31 '18 at 7:21
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
add a comment |
$begingroup$
It's $$(x-a^3)(x+a^3)=1$$ and solve two systems:
$$x-a^3=1$$ and $$x+a^3=1$$ or
$$x-a^3=-1$$ and $$x+a^3=-1,$$
which gives the answer for $(x,a)$:
$${(1,0), (-1,0)}.$$
answered Dec 31 '18 at 7:21
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
It's $$(x-a^3)(x+a^3)=1$$ and solve two systems:
$$x-a^3=1$$ and $$x+a^3=1$$ or
$$x-a^3=-1$$ and $$x+a^3=-1,$$
which gives the answer for $(x,a)$:
$${(1,0), (-1,0)}.$$
answered Dec 31 '18 at 7:21
Michael RozenbergMichael Rozenberg
111k1897201
edited Dec 31 '18 at 7:30
answered Dec 31 '18 at 7:21
Michael RozenbergMichael Rozenberg
111k1897201
answered Dec 31 '18 at 7:21
Michael RozenbergMichael Rozenberg
111k1897201
answered Dec 31 '18 at 7:21
Michael RozenbergMichael Rozenberg
111k1897201
111k1897201
add a comment |
add a comment |
$begingroup$
$x^2$ and $a^6$ are both perfect squares. And the only two perfect squares that differ by $1$ are $0$ and $1$ (because for any $nge 0$, the next biggest square after $n^2$ is $n^2+2n+1$).
answered Dec 31 '18 at 7:33
TonyKTonyK
44.1k358137
$endgroup$
add a comment |
$begingroup$
$x^2$ and $a^6$ are both perfect squares. And the only two perfect squares that differ by $1$ are $0$ and $1$ (because for any $nge 0$, the next biggest square after $n^2$ is $n^2+2n+1$).
answered Dec 31 '18 at 7:33
TonyKTonyK
44.1k358137
$endgroup$
add a comment |
$begingroup$
$x^2$ and $a^6$ are both perfect squares. And the only two perfect squares that differ by $1$ are $0$ and $1$ (because for any $nge 0$, the next biggest square after $n^2$ is $n^2+2n+1$).
answered Dec 31 '18 at 7:33
TonyKTonyK
44.1k358137
$endgroup$
$x^2$ and $a^6$ are both perfect squares. And the only two perfect squares that differ by $1$ are $0$ and $1$ (because for any $nge 0$, the next biggest square after $n^2$ is $n^2+2n+1$).
answered Dec 31 '18 at 7:33
TonyKTonyK
44.1k358137
answered Dec 31 '18 at 7:33
TonyKTonyK
44.1k358137
answered Dec 31 '18 at 7:33
TonyKTonyK
44.1k358137
answered Dec 31 '18 at 7:33
TonyKTonyK
44.1k358137
44.1k358137
add a comment |
add a comment |
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