Let a and n be positive integers, and a divide by 2n ^ 2. Try to explain that n ^ 2 + A is not an answer before the square number,

Another method:
Let n ^ 2 + a = m ^ 2, (2n ^ 2) / a = k, then a = 2 * n ^ 2 / K
Bring in
n^2+a=n^2+2*n^2/k=n^2*(1+2/k)
To make it a perfect square, 1 + 2 / K must be a perfect square
But when k = 1, 1 + 2 / k = 3 is not
When k = 2, 1 + 2 / k = 2 is not
k> When = 3, 1 + 2 / K is a fraction, not a fraction
So we can see the conclusion
As for your initial solution, the purpose of introducing K is to simplify a in n ^ 2 + a with N and K. if it is reduced to (4N ^ 2 + AK ^ 2) / (2k), it will make the problem more complicated and not achieve the purpose of simplification. When k = 1, 1 + 2 / k = 3 is not
When k = 2, 1 + 2 / k = 2 is not
k> When = 3, 1 + 2 / K is a fraction, not a fraction

It is proved that when n is a positive integer, the number represented by the cubic power of N + 3 times (the square of n) + 2n must be divisible by 3

(n + 2) + (n + 1) times the square of (n + 2)
There must be a multiple of 3, so the number represented by the cubic power of N + 3 times (the square of n) + 2n must be divisible by 3

Let a and n be positive integers, and a divide the square of 2n. Try to show that the square of N + A is not a square number
Come on. Okay, bonus points

a|2n^2
If n ^ 2 + a = B ^ 2
Then 2B ^ 2 = 2n ^ 2 + 2A
So a|2b ^ 2
If a is odd, then a | n ^ 2, a | B ^ 2
If a is even, then a / 2 | n ^ 2, a / 2 | B ^ 2
There are always c | n ^ 2 C | B ^ 2 a = C or 2C
Let C = x ^ 2 * y without square factor
Let y | (n / x) ^ 2 = y ^ 2 * n ^ 2, y | (B / x) ^ 2 = y ^ 2 * B ^ 2, then n = xyn, B = xyb
In duplicate:
Then B ^ 2Y = n ^ 2Y + 1 or n ^ 2Y + 2
1 = y (b ^ 2-N ^ 2) ---- obviously no solution
perhaps
2 = y (b ^ 2-N ^ 2) ---- obviously no solution
So. It can't be a perfect square
The proof is complete

Let a and n be positive integers, and a divide by 2n ^ 2. Try to explain that n ^ 2 + A is not an answer before the square number,