Find integer solutions for $m^3-n^2=2$
1
$begingroup$
Find integer solutions for $m^3-n^2=2$
I have seen multiple solutions like this one.
However, these have been complicated. I tried taking different modulos but could not achieve anything. My latest attempt included this:
$$m^3+1=n^2-1$$
$$(m+1)(m^2-m+1)=(n-1)(n+1)$$
But this does not help. Is there any solution that is easy. Thank you in advance.
elementary-number-theory
asked Dec 23 '18 at 7:42
user587054user587054
56511
$endgroup$
|
show 3 more comments
1
$begingroup$
Find integer solutions for $m^3-n^2=2$
I have seen multiple solutions like this one.
However, these have been complicated. I tried taking different modulos but could not achieve anything. My latest attempt included this:
$$m^3+1=n^2-1$$
$$(m+1)(m^2-m+1)=(n-1)(n+1)$$
But this does not help. Is there any solution that is easy. Thank you in advance.
elementary-number-theory
asked Dec 23 '18 at 7:42
user587054user587054
56511
$endgroup$
1
$begingroup$
For what it's worth, the equation $$y^2 = x^3 - 2$$ defines an elliptic curve, with the two lattice points $(3, 5)$ and $(3, -5)$.
$endgroup$
– Jose Arnaldo Bebita Dris
Dec 23 '18 at 7:46
2
$begingroup$
BTW, the equations $$m^3 - n^2 = 2$$ and $$m^3 + 1 = n^2 - 1$$ do not have the same solutions.
$endgroup$
– Jose Arnaldo Bebita Dris
Dec 23 '18 at 7:51
2
$begingroup$
The method in the linked thread is from several textbooks on number theory. The reason is probably that it is the simplest solution known to mankind. Modular arithmetic is unlikely to do much good exactly because we have that solution $m=3,n=5$. With respect to any modulus there will be solutions congruent to that. Some kind of a descent might still be possible (compare to the standard proof of irrationality of $sqrt2$), but if one existed, surely the number theory books would use that instead.
$endgroup$
– Jyrki Lahtonen
Dec 23 '18 at 7:52
$begingroup$
You might ask to simply explain that solution instead.
$endgroup$
– tarit goswami
Dec 23 '18 at 7:56
3
$begingroup$
@taritgoswami: I refer you to this MathOverflow Q&A, and I quote: "Finding all the integral points on an elliptic curve is a non-trivial computational problem."
$endgroup$
– Jose Arnaldo Bebita Dris
Dec 23 '18 at 8:05
|
show 3 more comments
1
1
1
1
$begingroup$
Find integer solutions for $m^3-n^2=2$
I have seen multiple solutions like this one.
However, these have been complicated. I tried taking different modulos but could not achieve anything. My latest attempt included this:
$$m^3+1=n^2-1$$
$$(m+1)(m^2-m+1)=(n-1)(n+1)$$
But this does not help. Is there any solution that is easy. Thank you in advance.
elementary-number-theory
asked Dec 23 '18 at 7:42
user587054user587054
56511
$endgroup$
Find integer solutions for $m^3-n^2=2$
I have seen multiple solutions like this one.
However, these have been complicated. I tried taking different modulos but could not achieve anything. My latest attempt included this:
$$m^3+1=n^2-1$$
$$(m+1)(m^2-m+1)=(n-1)(n+1)$$
But this does not help. Is there any solution that is easy. Thank you in advance.
elementary-number-theory
elementary-number-theory
asked Dec 23 '18 at 7:42
user587054user587054
56511
asked Dec 23 '18 at 7:42
user587054user587054
56511
edited Dec 23 '18 at 7:48
user587054
asked Dec 23 '18 at 7:42
user587054user587054
56511
asked Dec 23 '18 at 7:42
user587054user587054
56511
asked Dec 23 '18 at 7:42
user587054user587054
56511
56511
1
$begingroup$
For what it's worth, the equation $$y^2 = x^3 - 2$$ defines an elliptic curve, with the two lattice points $(3, 5)$ and $(3, -5)$.
$endgroup$
– Jose Arnaldo Bebita Dris
Dec 23 '18 at 7:46
2
$begingroup$
BTW, the equations $$m^3 - n^2 = 2$$ and $$m^3 + 1 = n^2 - 1$$ do not have the same solutions.
$endgroup$
– Jose Arnaldo Bebita Dris
Dec 23 '18 at 7:51
2
$begingroup$
The method in the linked thread is from several textbooks on number theory. The reason is probably that it is the simplest solution known to mankind. Modular arithmetic is unlikely to do much good exactly because we have that solution $m=3,n=5$. With respect to any modulus there will be solutions congruent to that. Some kind of a descent might still be possible (compare to the standard proof of irrationality of $sqrt2$), but if one existed, surely the number theory books would use that instead.
$endgroup$
– Jyrki Lahtonen
Dec 23 '18 at 7:52
$begingroup$
You might ask to simply explain that solution instead.
$endgroup$
– tarit goswami
Dec 23 '18 at 7:56
3
$begingroup$
@taritgoswami: I refer you to this MathOverflow Q&A, and I quote: "Finding all the integral points on an elliptic curve is a non-trivial computational problem."
$endgroup$
– Jose Arnaldo Bebita Dris
Dec 23 '18 at 8:05
|
show 3 more comments
1
$begingroup$
For what it's worth, the equation $$y^2 = x^3 - 2$$ defines an elliptic curve, with the two lattice points $(3, 5)$ and $(3, -5)$.
$endgroup$
– Jose Arnaldo Bebita Dris
Dec 23 '18 at 7:46
2
$begingroup$
BTW, the equations $$m^3 - n^2 = 2$$ and $$m^3 + 1 = n^2 - 1$$ do not have the same solutions.
$endgroup$
– Jose Arnaldo Bebita Dris
Dec 23 '18 at 7:51
2
$begingroup$
The method in the linked thread is from several textbooks on number theory. The reason is probably that it is the simplest solution known to mankind. Modular arithmetic is unlikely to do much good exactly because we have that solution $m=3,n=5$. With respect to any modulus there will be solutions congruent to that. Some kind of a descent might still be possible (compare to the standard proof of irrationality of $sqrt2$), but if one existed, surely the number theory books would use that instead.
$endgroup$
– Jyrki Lahtonen
Dec 23 '18 at 7:52
$begingroup$
You might ask to simply explain that solution instead.
$endgroup$
– tarit goswami
Dec 23 '18 at 7:56
3
$begingroup$
@taritgoswami: I refer you to this MathOverflow Q&A, and I quote: "Finding all the integral points on an elliptic curve is a non-trivial computational problem."
$endgroup$
– Jose Arnaldo Bebita Dris
Dec 23 '18 at 8:05
1
1
$begingroup$
For what it's worth, the equation $$y^2 = x^3 - 2$$ defines an elliptic curve, with the two lattice points $(3, 5)$ and $(3, -5)$.
$endgroup$
– Jose Arnaldo Bebita Dris
Dec 23 '18 at 7:46
$begingroup$
For what it's worth, the equation $$y^2 = x^3 - 2$$ defines an elliptic curve, with the two lattice points $(3, 5)$ and $(3, -5)$.
$endgroup$
– Jose Arnaldo Bebita Dris
Dec 23 '18 at 7:46
2
2
$begingroup$
BTW, the equations $$m^3 - n^2 = 2$$ and $$m^3 + 1 = n^2 - 1$$ do not have the same solutions.
$endgroup$
– Jose Arnaldo Bebita Dris
Dec 23 '18 at 7:51
$begingroup$
BTW, the equations $$m^3 - n^2 = 2$$ and $$m^3 + 1 = n^2 - 1$$ do not have the same solutions.
$endgroup$
– Jose Arnaldo Bebita Dris
Dec 23 '18 at 7:51
2
2
$begingroup$
The method in the linked thread is from several textbooks on number theory. The reason is probably that it is the simplest solution known to mankind. Modular arithmetic is unlikely to do much good exactly because we have that solution $m=3,n=5$. With respect to any modulus there will be solutions congruent to that. Some kind of a descent might still be possible (compare to the standard proof of irrationality of $sqrt2$), but if one existed, surely the number theory books would use that instead.
$endgroup$
– Jyrki Lahtonen
Dec 23 '18 at 7:52
$begingroup$
The method in the linked thread is from several textbooks on number theory. The reason is probably that it is the simplest solution known to mankind. Modular arithmetic is unlikely to do much good exactly because we have that solution $m=3,n=5$. With respect to any modulus there will be solutions congruent to that. Some kind of a descent might still be possible (compare to the standard proof of irrationality of $sqrt2$), but if one existed, surely the number theory books would use that instead.
$endgroup$
– Jyrki Lahtonen
Dec 23 '18 at 7:52
$begingroup$
You might ask to simply explain that solution instead.
$endgroup$
– tarit goswami
Dec 23 '18 at 7:56
$begingroup$
You might ask to simply explain that solution instead.
$endgroup$
– tarit goswami
Dec 23 '18 at 7:56
3
3
$begingroup$
@taritgoswami: I refer you to this MathOverflow Q&A, and I quote: "Finding all the integral points on an elliptic curve is a non-trivial computational problem."
$endgroup$
– Jose Arnaldo Bebita Dris
Dec 23 '18 at 8:05
$begingroup$
@taritgoswami: I refer you to this MathOverflow Q&A, and I quote: "Finding all the integral points on an elliptic curve is a non-trivial computational problem."
$endgroup$
– Jose Arnaldo Bebita Dris
Dec 23 '18 at 8:05
|
show 3 more comments
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1
$begingroup$
For what it's worth, the equation $$y^2 = x^3 - 2$$ defines an elliptic curve, with the two lattice points $(3, 5)$ and $(3, -5)$.
$endgroup$
– Jose Arnaldo Bebita Dris
Dec 23 '18 at 7:46
2
$begingroup$
BTW, the equations $$m^3 - n^2 = 2$$ and $$m^3 + 1 = n^2 - 1$$ do not have the same solutions.
$endgroup$
– Jose Arnaldo Bebita Dris
Dec 23 '18 at 7:51
2
$begingroup$
The method in the linked thread is from several textbooks on number theory. The reason is probably that it is the simplest solution known to mankind. Modular arithmetic is unlikely to do much good exactly because we have that solution $m=3,n=5$. With respect to any modulus there will be solutions congruent to that. Some kind of a descent might still be possible (compare to the standard proof of irrationality of $sqrt2$), but if one existed, surely the number theory books would use that instead.
$endgroup$
– Jyrki Lahtonen
Dec 23 '18 at 7:52
$begingroup$
You might ask to simply explain that solution instead.
$endgroup$
– tarit goswami
Dec 23 '18 at 7:56
3
$begingroup$
@taritgoswami: I refer you to this MathOverflow Q&A, and I quote: "Finding all the integral points on an elliptic curve is a non-trivial computational problem."
$endgroup$
– Jose Arnaldo Bebita Dris
Dec 23 '18 at 8:05