Question

Find the sum of the absolute values of all integral values of X that make the following expression prime:

(x^2-5x)^2 - 34

Answer 1

investigation:
x=-2 , (x^2 - 5x)^2 - 34 = 162
x=-1 , (x^2 - 5x)^2 - 34 = 2
x=0 , (x^2 - 5x)^2 - 34 = -34
x=1 , (x^2 - 5x)^2 - 34 = -18
x=2 , (x^2 - 5x)^2 - 34 = 2
x=3 , (x^2 - 5x)^2 - 34 = 2
x=4 , (x^2 - 5x)^2 - 34 = -18
x=5 , (x^2 - 5x)^2 - 34 = -34
x=6 , (x^2 - 5x)^2 - 34 = 2
x=7 , (x^2 - 5x)^2 - 34 = 162
x=8, (x^2 - 5x)^2 - 34 = 542
x=9 (x^2 - 5x)^2 - 34 = big

x>9 expression gets bigger and bigger BUT ALWAYS EVEN
if x is even:
x^2 is even and 5x is even,
the difference between 2 evens is even, squaring an even keeps it even.
so (x^2 - 5x)^2 is even, and subtacting 34 keeps it EVEN
if x is odd,
x^2 is odd and 5x is odd
the difference between two odds is EVEN, so again the whole expression is EVEN

The only prime number is 2, and we got a result of 2 when
x = -1 , 2, 3, and 6
The sum of the absolute values of those x's is 12