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Let $n\in \mathbb{N}$, if $4n^2-1=3m^2$ has a positive integer solution, show that $2n-1$ is a perfect square. For example $n=1$, it is clear.

and $n=13$ because $$ 4\cdot 13^2-1=3\cdot 15^2$$ and $$2n-1=25=5^2$$

Lisbeth
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math110
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3 Answers3

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$a^2-1=3b^2$ is a Pell equation, $a^2-3b^2=1$. The solutions correspond to $$ a_k + b_k \sqrt3 = (2+\sqrt3)^{k} $$ We have $$ a_{k+2}=4a_{k+1}-a_k , \quad a_0 = 1 , \quad a_1 = 2 $$ because $2+\sqrt3$ is a root of $x^2 - 4 x + 1$. Therefore $$ a_k \bmod 2 = 1,0,1,0,\dots \\ a_k \bmod 3 = 1,2,1,2,\dots $$

Write $4n^2-1=3m^2$ as $(2n)^2-3m^2=1$. Therefore, $2n=a_{2k+1}$ and so $2n \equiv 2 \bmod 3$.

Write $4n^2-1=3m^2$ as $(2n-1)(2n+1)=3m^2$. Since $3$ divides $2n+1$ and $2n-1$ and $2n+1$ are coprime, we must have that $2n-1$ is square.

lhf
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    to be explicit,$$\sqrt{2n-1}=\sqrt{a_{2k+1}-1}=\sqrt{\frac12(a_{2k}+1)}+\sqrt{\frac32(a_{2k}-1)}$$ – J. W. Tanner Sep 11 '19 at 00:46
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    I was trying to figure out if the question could be answered without invoking the solution to Pell's equation. Basically the last paragraph of this answer without the rest. $4n^2-1=3m^2\implies m$ is odd, so $3m^2+1\equiv 4\pmod 8$ so $n$ is odd so $\gcd(2n+1,2n-1)=1$; $3$ divides $2n-1$ or $2n+1$ but I couldn't see how to prove $3$ divides $2n+1$ and not $2n-1$ – J. W. Tanner Sep 11 '19 at 03:22
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    @J.W.Tanner, me too! – lhf Sep 11 '19 at 10:30
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Here I prove an explicit expression for $\sqrt{2n-1}: \sqrt{\frac12(4n-3m+1)}+\sqrt{\frac32(4n-3m-1)}$

From lhf's answer, $a_{2k+1}+b_{2k+1}\sqrt3=(a_{2k}+b_{2k}\sqrt3)(2+\sqrt3)=(2a_{2k}+3b_{2k})+(a_{2k}+2b_{2k})\sqrt3$,

so $a_{2k+1}=2a_{2k}+3b_{2k}$ and $b_{2k+1}=a_{2k}+2b_{2k}$ and therefore $a_{2k}=2a_{2k+1}-3b_{2k+1}$.

Also, from $a_{2k}^2-3b_{2k}^2=1$ we have $\sqrt{3(a_{2k}^2-1)}=3b_{2k}$.

Therefore, $\left(\sqrt{\frac12(a_{2k}+1)}+\sqrt{\frac32(a_{2k}-1)}\right)^2=\frac12(a_{2k}+1)+\frac32(a_{2k}-1)+\sqrt{3(a_{2k}^2-1)}$

$=2a_{2k}+b_{2k}-1=a_{2k+1}-1=2n-1.$

Note that $\sqrt{\frac12(a_{2k}+1)}$ and $\sqrt{\frac32(a_{2k}-1)}$ are integers,

by arguments similar to lhf's that $\sqrt{\dfrac{2n+1}3}$ and $\sqrt{2n-1}$ are.

J. W. Tanner
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An Experimental approach:

There are infinitely many integers like n such that:

$2n-1=k^2$; $k∈N$

They make following series:

$n=13, 41, 61, 181, 221 . . .$

therefore we may write:

$4\times 13^2-1=3\times 15^2$

$4\times 41^2-1=3\times 47.34^2$

$4\times 61^2-1=3\times 70.4^2$

$4\times 181^2-1=3\times 209^2$

$4\times 221^2-1=3\times 255.19^2$

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For $4 . n^2-1=3 .m^2$ The condition $2n-1=k^2$ is necessary but not sufficient. That is if equation ($4 . n^2-1=3 .m^2$) has integer solution then ($2n-1$) is definitely is a square.

sirous
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