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[tex]2{\leq}n{\leq}80[/tex]

For how many values the expression [tex]\frac{(n+1)n(n-1)}{8}[/tex] takes positive and integer values?

I solved it that way...

[tex]\frac{(n+1)n(n-1)}{8}=\frac{(n^{2}-1)n}{8}[/tex]

(n^2 - 1)n must have 8 as one of its factor.

Either n is a multiple of 8, or n^2 - 1 is. Also the case were n^2 - 1 has 4 as one of its factors, n having 2, and vice-versa, is impossible - if n^2 - 1 is even, n is odd, and vice-versa.

So every mutliple of 8 up to 80 is a possible value. So there is 10 values.

Let's list those numbers

8, 16 , 24 , 32 , 40 , 48 , 56 , 64 , 72, 80

Add one to each one of these values

9, 17, 25, 33, 41, 49, 65 , 73, 81

There is 4 perfect square in this list. So if n^2 - 1 is a multiple of 8, then there is 4 possible values for n.

10 + 4 = 14

So 14 possibilities in total. But the true awnser is not what I found. What is wrong in my reasoning?