Find all prime numbers P and positive integers m satisfying 2P2 + P + 8 = m2-2m

Find all prime numbers P and positive integers m satisfying 2P2 + P + 8 = m2-2m

Let P (2P + 1) = (M-4) (M + 2), because P is a prime, so p is a factor of (M-4), or P is a factor of (M + 2). (5 points) (1) if P divides (M-4), let M-4 = KP, K is a positive integer, then M + 2 > KP, 3P2 > P (2P + 1) = (M-4) (M + 2) > k2p2, so K2 < 3, so k = 1, so m − 4 = PM + 2 = 2p + 1, the solution is p = 5m = 9. (10 points) (2) if P divides (M-4) = PM + 2 = 2p + 1, the solution is p = 5m = 9+ 2) When p > 5, there are M-4 = kp-6 > kp-p = P (k-1), 3P2 > P (2P + 1) = (M-4) (M + 2) > k (k-1) P2, so K (k-1) < 3, so k = 1, or 2, because P (2P + 1) = (M-4) (M + 2) is odd, so K ≠ 2, so k = 1, so m − 4 = 2p + 1m + 2 = P, which is impossible. When p = 5, m2-2m = 63, M = 9; when p = 3, m2-2m = 29, there is no positive integer When p = 2, m2-2m = 18, there is no positive integer solution

Find all prime numbers P and integers m satisfying square of 2p + P + 8 = square of m-2

Is the title wrong?
It should be 2p ^ 2 + P + 8 = m ^ 2-2m. We know that P is prime and M is a positive integer
It's like just one set of solutions
p(2p+1)=m^2-2m-8=(m-4)(m+2)
p=m-4
2p+1=m+2
Solution
p=5
m=9

M is a positive integer, M & # 178; + m + 7 is a complete square number

Let: M & # 178; + m + 7 = (M + a) &# 178; (M is a positive integer, a is an integer)
m²+m+7=m²+2am+a²
∴（2a-1）m=7-a²
When m = 1, a = 6
When a = 2, M = 1
When a > = 3, M