Given that m and N are positive integers, the square of M = the square of N + 45, find the value of M and n

Given that m and N are positive integers, the square of M = the square of N + 45, find the value of M and n

m²=n²+45
m²-n²=45
(m+n)(m-n)=45
M + n = 45, M-N = 1 or M + n = 9, M-N = 5 or M + n = 15, M-N = 3
Solution
m=23,n=22
or
m=7,n=2
or
m=9,n=6

It is known that m and N are positive integers, and Mn │ M & # 178; + n & # 178; + m. It is proved that M is a complete square number

Mn │ (m ^ 2 + n ^ 2 + m), that is, m | n ^ 2
N | (m ^ 2 + m) -- > n | m (M + 1), because m, M + 1 are coprime, so we need: M = kn or M + 1 = kn
When m + 1 = kn, M = kn-1. Because kn-1 and N are coprime, it is impossible for m | n ^ 2, so m + 1 cannot be kn
When m = kn, K | n is obtained from m | n ^ 2, that is, n = Kr, so m = k ^ 2R
mn=k^3r^2
m^2+n^2+m=k^2r(k^2r+r+1)
So from Mn | (m ^ 2 + n ^ 2 + 1), we get Kr | (k ^ 2R + R + 1), so R | (1), so r = 1
Therefore, M = k ^ 2 is a complete square number

It is known that m and N are positive integers, and Mn │ m ∧ 2 + n ∧ 2 + m. It is proved that M is a complete square number

Mn │ (m ^ 2 + n ^ 2 + m), that is, m | n ^ 2
N | (m ^ 2 + m) -- > n | m (M + 1), because m, M + 1 are coprime, so we need: M = kn or M + 1 = kn
When m + 1 = kn, M = kn-1. Because kn-1 and N are coprime, it is impossible for m | n ^ 2, so m + 1 cannot be kn
When m = kn, K | n is obtained from m | n ^ 2, that is, n = Kr, so m = k ^ 2R
mn=k^3r^2
m^2+n^2+m=k^2r(k^2r+r+1)
So from Mn | (m ^ 2 + n ^ 2 + 1), we get Kr | (k ^ 2R + R + 1), so R | (1), so r = 1
Therefore, M = k ^ 2 is a complete square number