$existsinfty$ many pairs of consecutive squares s.t. their sum is also a square












1












$begingroup$


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?










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  • 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


















1












$begingroup$


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?










share|cite|improve this question











$endgroup$








  • 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
















1












1








1





$begingroup$


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?










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 13:55







Algebear

















asked Dec 16 '18 at 13:43









AlgebearAlgebear

655319




655319








  • 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




    $begingroup$
    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












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