Get rid of the dumb computer programs and use smart inequalities to limit the solution space!
Let m*(m+1) be the product. Multiply both sides by 4 and add 1. Then 4n^4-32n+61 = (2m+1)^2.
This is an odd square, so I will bound it between a couple of odd squares:
(2n^2-5) < 4n^4-32n+61 < (2n^2+1)^2
There are two possible ways for a solution to occur, either a square within the bounds or one of the inequalities fail.
There are two odd squares within the the range: (2n^2-3)^2 and (2n^2-1)^2
If (2n^2-3)^2 = 4n^4-32n+61, then 4n^2-8n+13=0, but this has no integer solutions.
If (2n^2-1)^2 = 4n^4-32n+61, then n^2-8n+15=0. Then n=3 and n=5 are potential solutions.
Checking if there are n which fail the inequality:
If (2n^2-5) >= 4n^4-32n+61, then (5n-4)^2+26>0, this is true for all integer n, so no failure cases here.
4n^4-32n+61 >= (2n^2+1)^2, then n^2-8n+15=(n-3)*(n-5)>=0. Then n=3, n=4, and n=5 are potential solutions.
So all the possible n are n=3, n=4, and n=5.
If n=3 then n^4-8n+15=72=8*9, a solution.
If n=4 then n^4-8n+15=239, not a solution.
If n=5 then n^4-8n+15=600=24*25, a solution.
All integer solutions are the two values n=3 and n=5.
Solution
