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mn+3m-8n=59 -> mn+3m=8n+59 -> m*(n+3)=8n+24+35-> (n+3)*(m-8)=35
1. 1*35 option doesn't work as n+3 must be >3,
2. 35*1 --> m=9, n=32,
3. 5*7 --> m=15, n=2,
4. 7*5 --> m=13, n=4
Total 3 pairs
What is the least positive integer n that can be placed in the following expression:
n!(n+1)!(2n+1)! - 1
and yields a number ending with thirty digits of 9's.
If 1^3 + 2^3 + 3^3 = m^2 where m is also an integer.
What are the next three consecutive positive integers such that the sum of their individual cubes is equal to a perfect square?
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