Q13/128) Consider n^2 + 20n + k = m^2 where k is an integer. As k varies, you can get cases of 0 or more solutions. What values of k causes 0 solutions?

if k is 2 mod 4 then there is no solution

Reason:
(n^2+20n + k) = (n+10)^2 + (k-100)

Now if n is odd then (n+10)^2 + (k-100) mod 4 = 1 + 2 = 3 so cannot be perfect square
if n is even then (n+10)^2 + (k-100) mod 4 = 2 so cannot be perfect square

so n is neither odd nor even .

so no solution