$existsinfty$ many pairs of consecutive squares s.t. their sum is also a square
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First of all, the term "pairs" is two of them, I assume (question's formulation is rather difficult to understand for me). So I guess this is the statement:
$$existsinftytext{ many pairs of consecutive squares }x^2, (x+1)^2text{ s.t. }x^2+(x+1)^2=y^2 x,yinmathbb{N}.$$
So there are two theorems I know of that might be of any use here:
1. An odd prime $p$ can be written as sum of squares iff $pequiv1$ mod $4$.
2. $ninmathbb{N}$ is the sum of squares iff primes that are $3$ mod $4$ occur an even number of times in the prime factorisation of $n$.
We can work out the LHS like so: $x^2+(x+1)^2=2x(x+1)+1$. We know that either $x$ or $x+1$ must be even; hence, the expression is of the form $4k+1$ with $kinmathbb{Z}$. From this we can conclude that the sum of two consecutive squares is always congruent $1$ modulo $4$ which also implies that primes that are $pequiv3$ mod $4$ that divide the sum of the consecutive squares must occur an even number of times. To this point, this does not prove anything significant, I think.
Does anyone have a hint on how to prove the statement?
number-theory modular-arithmetic sums-of-squares
asked Dec 16 '18 at 13:43
AlgebearAlgebear
628319
$endgroup$
add a comment |
$begingroup$
First of all, the term "pairs" is two of them, I assume (question's formulation is rather difficult to understand for me). So I guess this is the statement:
$$existsinftytext{ many pairs of consecutive squares }x^2, (x+1)^2text{ s.t. }x^2+(x+1)^2=y^2 x,yinmathbb{N}.$$
So there are two theorems I know of that might be of any use here:
1. An odd prime $p$ can be written as sum of squares iff $pequiv1$ mod $4$.
2. $ninmathbb{N}$ is the sum of squares iff primes that are $3$ mod $4$ occur an even number of times in the prime factorisation of $n$.
We can work out the LHS like so: $x^2+(x+1)^2=2x(x+1)+1$. We know that either $x$ or $x+1$ must be even; hence, the expression is of the form $4k+1$ with $kinmathbb{Z}$. From this we can conclude that the sum of two consecutive squares is always congruent $1$ modulo $4$ which also implies that primes that are $pequiv3$ mod $4$ that divide the sum of the consecutive squares must occur an even number of times. To this point, this does not prove anything significant, I think.
Does anyone have a hint on how to prove the statement?
number-theory modular-arithmetic sums-of-squares
asked Dec 16 '18 at 13:43
AlgebearAlgebear
628319
$endgroup$
3
$begingroup$
You want $$2x^2+2x+1=y^2iff 4x^2+4x+1+1=2y^2iff (2x+1)^2+1=2y^2$$ and this is a Pell equation of the form $z^2-2y^2=-1$.
$endgroup$
– Galc127
Dec 16 '18 at 14:02
3
$begingroup$
The list of values for $y$ is OEIS A$001653$. That link has multiple descriptions, including connections to Pell's Equation.
$endgroup$
– lulu
Dec 16 '18 at 14:03
add a comment |
$begingroup$
First of all, the term "pairs" is two of them, I assume (question's formulation is rather difficult to understand for me). So I guess this is the statement:
$$existsinftytext{ many pairs of consecutive squares }x^2, (x+1)^2text{ s.t. }x^2+(x+1)^2=y^2 x,yinmathbb{N}.$$
So there are two theorems I know of that might be of any use here:
1. An odd prime $p$ can be written as sum of squares iff $pequiv1$ mod $4$.
2. $ninmathbb{N}$ is the sum of squares iff primes that are $3$ mod $4$ occur an even number of times in the prime factorisation of $n$.
We can work out the LHS like so: $x^2+(x+1)^2=2x(x+1)+1$. We know that either $x$ or $x+1$ must be even; hence, the expression is of the form $4k+1$ with $kinmathbb{Z}$. From this we can conclude that the sum of two consecutive squares is always congruent $1$ modulo $4$ which also implies that primes that are $pequiv3$ mod $4$ that divide the sum of the consecutive squares must occur an even number of times. To this point, this does not prove anything significant, I think.
Does anyone have a hint on how to prove the statement?
number-theory modular-arithmetic sums-of-squares
asked Dec 16 '18 at 13:43
AlgebearAlgebear
628319
$endgroup$
First of all, the term "pairs" is two of them, I assume (question's formulation is rather difficult to understand for me). So I guess this is the statement:
$$existsinftytext{ many pairs of consecutive squares }x^2, (x+1)^2text{ s.t. }x^2+(x+1)^2=y^2 x,yinmathbb{N}.$$
So there are two theorems I know of that might be of any use here:
1. An odd prime $p$ can be written as sum of squares iff $pequiv1$ mod $4$.
2. $ninmathbb{N}$ is the sum of squares iff primes that are $3$ mod $4$ occur an even number of times in the prime factorisation of $n$.
We can work out the LHS like so: $x^2+(x+1)^2=2x(x+1)+1$. We know that either $x$ or $x+1$ must be even; hence, the expression is of the form $4k+1$ with $kinmathbb{Z}$. From this we can conclude that the sum of two consecutive squares is always congruent $1$ modulo $4$ which also implies that primes that are $pequiv3$ mod $4$ that divide the sum of the consecutive squares must occur an even number of times. To this point, this does not prove anything significant, I think.
Does anyone have a hint on how to prove the statement?
number-theory modular-arithmetic sums-of-squares
number-theory modular-arithmetic sums-of-squares
asked Dec 16 '18 at 13:43
AlgebearAlgebear
628319
asked Dec 16 '18 at 13:43
AlgebearAlgebear
628319
edited Dec 16 '18 at 13:55
Algebear
asked Dec 16 '18 at 13:43
AlgebearAlgebear
628319
asked Dec 16 '18 at 13:43
AlgebearAlgebear
628319
asked Dec 16 '18 at 13:43
AlgebearAlgebear
628319
628319
3
$begingroup$
You want $$2x^2+2x+1=y^2iff 4x^2+4x+1+1=2y^2iff (2x+1)^2+1=2y^2$$ and this is a Pell equation of the form $z^2-2y^2=-1$.
$endgroup$
– Galc127
Dec 16 '18 at 14:02
3
$begingroup$
The list of values for $y$ is OEIS A$001653$. That link has multiple descriptions, including connections to Pell's Equation.
$endgroup$
– lulu
Dec 16 '18 at 14:03
add a comment |
3
$begingroup$
You want $$2x^2+2x+1=y^2iff 4x^2+4x+1+1=2y^2iff (2x+1)^2+1=2y^2$$ and this is a Pell equation of the form $z^2-2y^2=-1$.
$endgroup$
– Galc127
Dec 16 '18 at 14:02
3
$begingroup$
The list of values for $y$ is OEIS A$001653$. That link has multiple descriptions, including connections to Pell's Equation.
$endgroup$
– lulu
Dec 16 '18 at 14:03
3
3
$begingroup$
You want $$2x^2+2x+1=y^2iff 4x^2+4x+1+1=2y^2iff (2x+1)^2+1=2y^2$$ and this is a Pell equation of the form $z^2-2y^2=-1$.
$endgroup$
– Galc127
Dec 16 '18 at 14:02
$begingroup$
You want $$2x^2+2x+1=y^2iff 4x^2+4x+1+1=2y^2iff (2x+1)^2+1=2y^2$$ and this is a Pell equation of the form $z^2-2y^2=-1$.
$endgroup$
– Galc127
Dec 16 '18 at 14:02
3
3
$begingroup$
The list of values for $y$ is OEIS A$001653$. That link has multiple descriptions, including connections to Pell's Equation.
$endgroup$
– lulu
Dec 16 '18 at 14:03
$begingroup$
The list of values for $y$ is OEIS A$001653$. That link has multiple descriptions, including connections to Pell's Equation.
$endgroup$
– lulu
Dec 16 '18 at 14:03
add a comment |
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$begingroup$
You want $$2x^2+2x+1=y^2iff 4x^2+4x+1+1=2y^2iff (2x+1)^2+1=2y^2$$ and this is a Pell equation of the form $z^2-2y^2=-1$.
$endgroup$
– Galc127
Dec 16 '18 at 14:02
3
$begingroup$
The list of values for $y$ is OEIS A$001653$. That link has multiple descriptions, including connections to Pell's Equation.
$endgroup$
– lulu
Dec 16 '18 at 14:03