Find ALL integer solutions (n,m) of the equation $(n+1)(2n+1)=10m^2$

Question

What I'm thinking is that since RHS contains a $10$, so at least one of the factors $(n+1)$ OR $(2n+1)$ must be divisible by 2. $(2n+1)$ is out of the question, so that means $(n+1)$ must be even, or $n$ must be odd.

If $(n+1)$ is divisible by $5$ also, then :

$n+1 = 5k_1 = 2k_2$ (for some ($k_1,k_2\in {\mathbb Z}$))

I do not know what to do with that $m^2$ .

Can anyone tell me how to proceed ?

EDIT : I found a factorized equivalent for the given statement : $(n+3m)(n-3m) + (n-m)(n+m) + (3n + 1) =0 $ . Is it something useful ?

Answer 1

Let $x=n+1$. The equation becomes $$x(2x-1)=10m^2 \\ 2x^2-x-10m^2=0 \\ x=\frac{1 \pm \sqrt{1+80m^2}}{4}$$

It follows that $1+80m^2$ must be a perfect square. Thus $$y^2=1+80m^2 \Rightarrow\\ y^2-80m^2=1$$

This is the Pell equation, and has solutions. Now, for each solution $(y,m)$, $y$ must be odd. Then, each of $1+y, 1-y$ is even, and their sum is $2$, meaning that exactly one of them is divisible by $4$ (look at them $\pmod{4}$).

That particular value leads to an integer value of $x$.

Answer 2

Since $n+1$ and $2n+1$ are coprime, they must be of the form $\pm k_1^2$ and $\pm 10k_2^2$ (option A), or $\pm 2k_1^2$ and $\pm 5k_2^2$ (option B). Let's explore both options in more detail.

(A) Since $2n+1$ is odd, it can't be divisible by 10, hence it equals $\pm k_1^2$, and $n+1=\pm10k_2^2$.
Positive branch (that is, interpreting both $\pm$ as +): $20k_2^2-1=k_1^2$. This thing can't have solutions modulo 4, hence it can't have solutions at all.
Negative branch: $-20k_2^2-1=-k_1^2$, so $20k_2^2+1=k_1^2$. This is a Pell equation, and it produces an infinite sequence of solutions: $(n,m)=(-1,0)$, then $(-41,18)$, then $(-12\,961,5\,796)$, then $(-4\,173\,161,1\,866\,294)$ and so on.

(B) Since $2n+1$ is odd, it equals $\pm5k_1^2$, and $n+1=\pm2k_2^2$.
Positive branch: $4k_2^2-1=5k_1^2$. This thing can't have solutions modulo 4 either, hence it can't have solutions at all.
Negative branch: $-4k_2^2-1=-5k_1^2$, so $4k_2^2+1=5k_1^2$. This is again a Pell equation (maybe generalized or something), and it produces another infinite sequence of solutions: $(n,m)=(-3,1)$, then $(-723,323)$, then $(-232\,563,104\,005)$ and so on.

Needless to say, you may put a minus before $m$ in any solution.

Answer 3

I got this. We have $(n+1)(2n+1)=2n^2+3n+1=10m^2$, if we multiply last equation by $8$ we get $16n^2+24n+8=80m^2$, but we can write this equation as $$(4n+3)^2-1=80m^2.$$

So if we set $4n+3=x$ we get the Pell-equation $x^2-80m^2=1$. The fundamental solution of this equation is $x=9$ and $m=1$, hence we have the following recursive formulas $$x_{r+1}=9x_r+80m_r,$$ $$m_{r+1}=x_r+9m_r,$$ where $r\in\Bbb{N}$. Therefore $$n=\frac{9x_r+80m_r-3}{4}=\frac{9x_r-3}{4}+20m_r.$$

This means that in order to have $n\in\Bbb{Z}$ we must have $9x_r\equiv 3\pmod 4$, equivalently $x_r\equiv 3\pmod 4$. Now, from $x_{r+1}=9x_r+80m_r$ it's easy to prove by induction that $x_r\equiv 1\pmod4$ for every $r\in\Bbb{N}$, so $-x_r\equiv 3\pmod 4$ and this lead us to infinitely many solutions, i.e. for every $x_r$ which is a solution of $x^2-80m^2=1$, we just need to take $-x_r$ in order to have $n\in\Bbb{Z}$. Hence all the solutions are $$(n,m)=\Bigl(\frac{-x_r-3}{4}, \pm m_r\Bigr).$$