Pell's equation:

$\begin{aligned} m^2 & = & 2n^2 + n \\ 4m^2 & = & 8n^2 + 4n \\ & = & 2 ( (2n+1)^2 - (2n+1)), \text{ let } p = 2n+1 \\ & = & 2p^2 -p \\ 16m^2 & = & 2 (4p^2 - 2p) \\ & = & 2 ( (2p-1)^2 - 1) \\ 2 & = & 2(2p-1)^2 - 16m^2 \\ 1 & = & (2p-1)^2 - 8m^2 \\ & = & (4n+1)^2 - 8(m^2) \\ \end{aligned}$

Knowing the two initial values $n_1 = 4 \Rightarrow m_1 = 6$ , we get the following pairs of $(n,m)$ less than $10^6$ : $(144, 204),( 4900, 6930),(166464, 235416)$

The second smallest $n$ is $\boxed{144}$

It reduces to n(2n+1)=m^2. now from here it is clear that n must be a perfect square. so it becomes a Pell's Equation that b^2- 2 .a^2=1.,whose initial solutions are (b,a)=(3,2) where n=a^2 and m/n=b. the other solutions are given by (p,q) where (3+2 /2)^2=p+q /2,hence the next immediate solution is 2.3.2=12. Thank You

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from math import *
nums=[]
for i in range(1,10**6+1):
    if sqrt((2*(i**2))+i).is_integer():
        nums.append(i)
print nums[1]

Python code

Here is a solution without Pell's using modular arithmetic.

It is clear $n=0, 3 \mod 4.$ We show $n=3 \mod 4$ fails. Let $n=4k+3$ and then $(4k+3)(8k+7)=m^2.$ Since $(4k+3)$ is not a perfect square this implies $gcd(4k+3, 8k+7)>1$ which by Euclid is false, contradiction. So let $n=4k$ and so $4k(8k+1).$ We have $k(8k+1)$ a perfect square. By Euclid $gcd(k, 8k+1)=1$ so $k$ is a perfect square. Simple testing gives us $k=1, k=36$ as the smallest working solutions. We see $n=4 \cdot 36 = 144.$

I did trial and error. :v

Multiply both sides of the equation by 8, so we can complete the square on the LHS (left hand side) using integer coefficients: $\begin{aligned} 2n^2+n&=m^2&&&\underset{\cdot 8}{\Leftrightarrow}&&&&(4n+1)^2 - 1 &= 2(2m)^2&&&\Leftrightarrow &&&&x^2-2y^2&=1&&&&\left|x:=4n+1>0,\quad y:=2m\geq0\right. \end{aligned}$ The vector $\vec{x}:=(x,\:y)^T$ satisfies Pell's equation, the non-negative solutions are $\vec{x}_k:=(x_k,\:y_k)^T$ . The trivial solution is $\vec{x}_0=(1;\:0)^T$ . Use continued fractions or simply guess the fundamental solution $\vec{x}_1=(3;\:2)^T$ to get all other positive solutions $\vec{x}_k$ in increasing order: $\begin{aligned} \vec{x}&=\begin{pmatrix} \pm x_k\\ \pm y_k \end{pmatrix},&&&&&&&\vec{x}_{k+1}&=\begin{pmatrix} 3&4\\ 2&3 \end{pmatrix}\vec{x}_k,&&&\vec{x}_0&=\begin{pmatrix}1\\0\end{pmatrix} \end{aligned}$ We don't care about non-positive solutions, so we only need to consider $\vec{x}=\vec{x}_k$ with $k>0$ . Of those, only the ones that satisfy the condition " $x_k\mod 4=1$ " lead to integer values for $n$ . In iteration $k=4$ we find the second smallest $n\in\mathbb{N}$ : $\begin{array}{r|r|r|r|r|r} k &1 & 2 & 3 & 4 &\cdots\\\hline x_k & 3 & 17 & 99 & 577 &\\ y_k & 2 & 12 & 70 & 408&\\\hline m & 1 & 6 & 35 & 204&\\ n & & 4 & & \boxed{144} & \end{array}$

Rem.: With the recurrence relation for $\vec{x}_k$ , it's possible to show that the relevant solutions with " $x_k\mod 4 = 1$ " are exactly the solutions with even $k=2K>0$ .

Assume that for a given prime $p$ , $n$ is written as $n=ap^t$ , where $p$ does not divide $a$ .

We have $2n^2+n=2a^2p^{2t}+ap^t=p^t(2ap^t+a)$ . We see that this expression is divisible by $p^t$ , but not by $p^{t+1}$ .

Since It is equal to the perfect square $m^2$ , $t$ must be even, and since this is the case for any $p$ , $n$ must be a perfect square.

We set $n=k^2$ .

We now have $2k^4+k^2=m^2$ so that $m$ must be a multiple of $k$

$2k^2+1=(\frac{m}{k})^2=u^2$ . We see that u is odd.

$2k^2=(u+1)(u-1)$ is a multiple of 4, so k must be even. Set $k= 2v$ to find

$8v^2+1=u^2$

Now try a few values for v and see if $8v^2+1$ is a perfect square:

Check: $2×144^2+144=41616=204^2$ So we find $n=\boxed{144}$

We could narrow the search further down by modular arithmetic: For example, if $v \equiv 2 \pmod{5}$ then $8v^2+1\equiv 3$ which is not in the set of squares $\{0,1,4\} \pmod{5}$ .

We find the conditions $v \in \{0,1,4\} \pmod{5}$ $v \in \{0,1,6\} \pmod{7}$ $v \in \{0,1,2,5,6,9,10\} \pmod {11}$ $v \in \{0,1,4,6,7,9,12\} \pmod {13}$ disqualifying over 90% of the candidates for $v$ .

Remaining candidates to check are then: $v \in \{1,6,35,104,126,134,155,160,176,181,189,204,...\}$ of which $8v^2+1$ is a square for $v \in \{1,6,35,204,...\}$

Remark: $0$ is not considered a natural number in this problem, otherwise $n=0$ would have been the smallest solution.

Here is a solution without using the information of 4 being the smallest $n$ ,

$2n^2 + n = m^2 \Rightarrow m = \sqrt{n}\sqrt{2n+1}$

This implies that $n$ and $2n+1$ are a perfect squares.

Since $2n+1$ is a perfect square, let $2n+1 = k^2, k \in \mathbb{Z}$ .

$k^2$ is obviously odd. Therefore, $k$ must be odd.

$2n + 1 = k^2 \Rightarrow 2n = (k+1)(k-1)$ .

Since $k$ is odd, $k+1$ and $k-1$ must be even. Therefore, $n$ must be even.

Since $n$ is an even perfect square, let $n = 4u^2, u \in \mathbb{Z}$ .

Therefore, $2n + 1 = k^2 \Rightarrow 2(4u^2) + 1 = k^2$

Finally we end up with a classic Pell's equation: $1 = k^2 - 8u^2$ .

Solving the Pell's equation, the three smallest solutions $(k,u)$ are $(1,0)$ , $(3,1)$ , $(17,6)$ .

Therefore, the three smallest values of $n$ which satisfy $2n^2 + n = m^2$ are $n=0$ (if you consider $0$ a perfect square), $n=4$ (given in the question), $n = \boxed{144}$ .

If you put m = n + k, you get:

n^2 - n(2k - 1) - k^2 = 0

The discriminant equals (2k - 1)^2 + 4k^2, which means

(2k - 1)^2 + (2k)^2 = E^2, where E is an integer.

The solution is easy to obtain using Euclid's formula for Pythagorean triplets (since (2k - 1) and 2k are consecutive integers), but I was way too lazy to do that so I just checked the list of the first few Pythagorean triples and found (119, 120, 169).

2k = 120, k = 60

n^2 - 119n - 3600 = 0

n = 144

include<iostream>

include<stdio.h>

include<math.h>

using namespace std; int main() { long a,q; float s; for(a=0;a<=1000;a++) { q=a+2*(pow(a,2)); s=sqrt(q); if(s==floor(s)) cout<<a<<"\n"; } return 0; }

n^2 + n^2 + n =n( 2 n +1 ) = m^2, so n is perfect square and n can divided by 2, so 2 n +1 is perfect square and m is can't divided by 2, due n is perfect square so 2 n +1 can be change to 2 a^2 +1 = b^2 , so (b+1)(b-1)= 2 a^2 ,so we get a is 2, 12 , etc so the smallest two is 12 and we get n is 144, thank you