If for some number $n\in \mathbb N$, the numbers $2n+1$ and $3n+1$ are perfect squares of integers, then prove that $8|n$.

if $2n+1=m^2$ and $3n+1=k^2$ then $k^2-m^2=3n-2n+1-1=n$ now I need to show that $8|k^2-m^2$ when you divide a some number $m^2$ with $8$ then remainder is $0,1,4$, so I need to show that $m^2$ and $k^2$ have the same remainder. But I do not know is this good way because I do not know how to prove that they must have the same remainder


If $k$ is odd, then $k^2\equiv1 \mod 8$. Hence $3n+1\equiv1\mod 8$, $2n+1\equiv 1\mod 8$, so $$(3n+1)-(2n+1)\equiv 1-1\equiv 0\mod 8$$

  • $\begingroup$ You need to prove that $k$ is odd to apply that, but you have not done so. $\endgroup$ – Bill Dubuque Nov 21 '18 at 20:32
  • $\begingroup$ Minor issue. How do we know $3n+1$ and $2n + 1$ are odd? $\endgroup$ – fleablood Nov 21 '18 at 20:49
  • $\begingroup$ Damned, you're right :-( Didn't see this one coming :-) $\endgroup$ – Nicolas FRANCOIS Nov 22 '18 at 20:32

If $2n+1=a^2$ and $3n+1=b^2$, then $a$ is odd. So $a=2k+1$ and $2n+1=(2k+1)^2=4k^2+4k+1$. Thus, $n=2k^2+2k=2k(k+1)$. This shows that $n$ is divisible by $4$ because $k(k+1)$ is even.

Now, $b^2=3n+1=6k(k+1)+1$. So, $b$ is also odd and $b=2l+1$. Then, $(2l+1)^2=6k(k+1)+1$ implies $2l(l+1)=3k(k+1)$. That is $k(k+1)$ is divisible by $4$ because $l(l+1)$ is even. Because $n=2k(k+1)$ and $k(k+1)$ is divisible by $4$, $n$ is divisible by $8$.

In fact, $3a^2-2b^2=3(2n+1)-2(3n+1)=1$. That is, $$(1+\sqrt{-2})(1-\sqrt{-2})a^2=1+2b^2=(1+\sqrt{-2}b)(1-\sqrt{-2}b).$$ Note that $\Bbb{Q}(\sqrt{-2})$ is a quadratic field with class number $1$, it is a ufd and we can talk about gcd. Because $\gcd(1+\sqrt{-2}b,1-\sqrt{-2}b)=1$, we get that either $$\frac{1+\sqrt{-2}b}{1+\sqrt{-2}}=(u+\sqrt{-2}v)^2$$ or $$\frac{1-\sqrt{-2}b}{1+\sqrt{-2}}=(u+\sqrt{-2}v)^2$$ for some $u,v\in\Bbb{Z}$. But up to sign switching $b\to -b$, we can assume that $$\frac{1+\sqrt{-2}b}{1+\sqrt{-2}}=(u+\sqrt{-2}v)^2=u^2-2v^2+2uv\sqrt{-2}.$$ That is, $$1+\sqrt{-2}b=u^2-2v^2-4uv+(u^2-2v^2+2uv)\sqrt{-2}.$$ So, $u^2-2v^2-4uv=1$ and so $(u-2v)^2-6v^2=1$. So, $(x,y)=(u-2v,v)$ is a solution to the Pell-type equation $$x^2-6y^2=1.$$ The solutions are known $$x+\sqrt{6}y=\pm(5+2\sqrt{6})^t$$ where $t\in\Bbb{Z}$. Since the sign switching $(u,v)\to(-u,-v)$ does not change anything, we can assume that $$u-2v+\sqrt{6}v=(5+2\sqrt{6})^t.$$ So, $$u-2v=\frac{(5+2\sqrt{6})^t+(5-2\sqrt{6})^{t}}{2}$$ and $$v=\frac{(5+2\sqrt{6})^t-(5-2\sqrt{6})^{t}}{2\sqrt{6}}.$$ That is, $$u=\frac{(2+\sqrt{6})(5+2\sqrt{6})^t-(2-\sqrt{6})(5-2\sqrt{6})^{t}}{2\sqrt{6}}.$$ That is, $$b=u^2-2v^2+2uv=\frac{(2+\sqrt{6})(5+2\sqrt{6})^{2t}+(2-\sqrt{6})(5-2\sqrt{6})^{2t}}{4}.$$ This gives $$a=\sqrt{\frac{1+2b^2}{3}}=u^2+2v^2=\frac{(3+\sqrt{6})(5+2\sqrt{6})^{2t}+(3-\sqrt{6})(5-2\sqrt{6})^{2t}}{6}.$$ So, we have $$n=\frac{(5+2\sqrt{6})^{4t+1}-10+(5-2\sqrt{6})^{4t+1}}{24},$$ where $t\in\mathbb{Z}$. So the first seven values of $n$ are $$n=0,40,3960, 388080,38027920,3726348120,365144087880.$$ That is, $n=n(t)$ satisfies $n(0)=0$ and $n(-1)=40$ with $$n(t-1)+n(t+1)=9602 n(t)+4000$$ for all $t\in\mathbb{Z}$. So, not only $8$ divides $n$, $40$ divides $n$ for every such $n$.

I think it is easier to re-parametrize $n$ using non-negative integers instead of all integers. Let $$n_t=\frac{(5+2\sqrt{6})^{2t+1}-10+(5-2\sqrt{6})^{2t+1}}{24},$$ for non-negative integer $t$. So, $n_0=0$, $n_1=40$, and $$n_{t+2}=98n_{t+1}-n_t+40.$$ We have the corresponding $a=a_t$ and $b=b_t$ to $n=n_t$: $a_0=1$, $a_1=9$, and $$a_{t+2}=10a_{t+1}-a_t,$$ as well as $b_0=1$, $b_1=11$, and $$b_{t+2}=10b_{t+1}-b_t.$$

$$ \begin{array}{ |c|c|c|c| } \hline t & n_t & a_t & b_t \\ \hline 0 & 0 & 1 &1 \\ 1 & 40 & 9& 11 \\ 2 & 3960 & 89 & 109 \\ 3 & 388080 & 881 &1079\\ 4 & 38027920 & 8721 & 10681 \\ 5 & 3726348120 & 86329 & 105731 \\ 6 & 365144087880 & 854569 & 1046629 \\\hline \end{array} $$

  • $\begingroup$ Does $4|n$ as well? $\endgroup$ – Yadati Kiran Nov 21 '18 at 16:59
  • $\begingroup$ If $8\mid n$, then $4\mid n$. $\endgroup$ – user614671 Nov 21 '18 at 17:00
  • $\begingroup$ I don't know what you did, but your answer was off by a factor of $2$. So, I fixed it. $\endgroup$ – Batominovski Nov 21 '18 at 18:31
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    $\begingroup$ I have a feeling that you can directly apply some results from Pell's equations, without having to deal with $\mathbb{Q}(\sqrt{-2})$. But I haven't thought about it well enough. $\endgroup$ – Batominovski Nov 21 '18 at 18:41
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    $\begingroup$ @Displayname I don't think Snookie was trying to prove $40\mid n$. The aim was probably to find all $n$. The result $40\mid n$ is a nice little consequence. And I think we can use Pell's equation $x^2-6y^2=-2$ or $x^2-6y^2=3$ without having to deal with $\mathbb{Q}(\sqrt{-2})$. However, I am not sure whether these Pell-type equations have unique fundamental solutions. $\endgroup$ – Batominovski Nov 21 '18 at 21:09

It's a standard exercise that if $b > a$ and $a$ and $b$ are both odd that

i) $a + b$ and $a-b$ are both even and that

ii) exactly one of $a+b$ and $a-b$ is divisible by $4$.

Pf: Let $b = 2m + 1$ and $a = 2n+1$ so $b-a = 2(m-n)$ and $b+a = 2(m+n+1)$ are both even.

If $m$ and $n$ are both even or both odd then $m-n$ is even and $m+n + 1$ is odd so $4|b-a$ but $4\not \mid b+a$. If $m$ and $n$ are opposite paritty then the exact opposite occurs; $m-n$ is odd and $m+n +1$ is even so $4\not \mid b-a$ and $4\mid b+a$.

As a consequence $b^2 - a^2 = (b-a)(b+a)$ is divisible by $8$.

So if we have $2n + 1 = a^2$ then $a^2$ is odd so $a$ is odd.

AND we have $2n = a^2 -1 = (a-1)(a+1)$ is divisible by $8$ so $n$ is even.

And if $3n + 1=b^2$ then we have $n$ is even so $b^2$ is odd so $b$ is odd.

So $n = (3n+1)-(2n+1) =b^2 - a^2$ where $a,b$ are both odd..... which is divisible by $8$.


Odd squares are all congruent to $1$ mod $8$, so if $2n+1$ is a square, we must have $4\mid n$. Writing $n=4n'$, we now have $3n+1=12n'+1$, which, if it's square, must also be congruent to $1$ mod $8$, so we must have $2\mid n'$. Writing $n'=2n''$, we have $n=4n'=8n''$, so $8\mid n$.


We have $2n+1, 3n+1 \equiv \{0,1,4 \} \mod 8,$ whereupon checking $n = 0, 1, \dots, 7,$ we see that only $n \equiv 0 \mod 8$ works.

Snookie has shown that $40 | n.$ Indeed, we can show $5 | n$ in a similar fashion by noting that $2n + 1 \equiv \{0, 1\} \mod 5$ implies $n \equiv 0 \mod 5$ by checking $0, 1, \dots, 4.$

Since $n=40$ works, this is the strongest result that we can prove.


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