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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 ?

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    Note that $n+1$ and $2n+1$ do not have any common factors, since $2(n+1) = 2n+2$ – 2017-01-12
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    Oh ! Didn't think about that .... But I'm still confused – 2017-01-12
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    The factorized form might have been useful, except you didn't find any. Also, what's with that $I$? Integer numbers are commonly denoted $\mathbb Z$. – 2017-01-12
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    Essentially, you need to solve $x^2 - 5 y^2 = \pm 1$. – 2017-01-12
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    So you will need to consider $m=rs$, where $\gcd(r,s)=1$, allocating factors to the two parts. – 2017-01-12

3 Answers 3

2

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 $x$.

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    I had a vague feeling that there is a simpler way of looking at this equation, which would combine both my series into one. Thank you for confirming that. – 2017-01-12
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    The discriminant is $1+80m^2$. – 2017-01-12
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    @Xam Ups, Fixed it, Thank you. – 2017-01-12
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    @N.S. no problem. Btw, since OP is not here, could you tell me what do you think about my solution? – 2017-01-12
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    @Xam I think that it is the same basic idea, which is not surprising since it is the natural one, but explained differently. You solved the quadratic equation by completing the square instead of quadratic formula, which makes the presentation a bit nicer. – 2017-01-12
2

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 $$(n,m)=\Bigl(\frac{-x_r-3}{4}, \pm m_r\Bigr).$$

1

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.