>>12479923The first one is pretty easy
Suppose n^2 + d = m^2, then there are three cases: when m-n is not a divisor of 2n^2, when m-n is a divisor of 2n^2, but not a divisor of 2n, and when m-n is a divisor of 2n.
In the first case, d clearly can’t be a divisor of 2n^2, so there we go. In the second case, m-n must contain a prime from n such that its multiplicity is greater than its multiplicity in n. So, the multiplicity of that prime in 2n + m-n must be at least its multiplicity in n. Therefore, d = (2n + m-n)(m-n) multiplicity of the prime must be strictly greater than its multiplicity in 2n^2, so d can’t be a divisor of 2n^2. Finally, in the third case, we get d = (2n/(m-n) + 1)(m-n)^2.
2n/(m-n) + 1 produces irrelevant factors for (2n/(m-n))^2 = 2*2n^2/(m-n)^2, so, d can’t be a divisor of 2n^2.
So, the equivalence n^2 + d = m^2, where d is a divisor of 2n^2 is bullshit.
