If x and y are positive integers and x>y, is (x-y)^3+(3y+1)^5 odd?

Hi VIVA1060 no problem!

Whatever you see here is not proper math of course, I am only reducing the exponents because it does not change if the terms are odd or even.

If x was even then \(x^2, x^3, x^4, x^5, .... , x^n\) are all even, where n is any positive integer.

If x was odd then \(x^2, x^3, x^4, x^5, .... , x^n\) are all odd, where n is any positive integer.

Thus we have this one-to-one relation we can conclude, on the other hand we know if \(x^n\) is even/odd, then \(x\) must be even/odd.

Now let's look at the problem again, \((x - y)^3\) can be reduced to \(x - y\) for the sake of this problem. If \((x - y)^3\) was even/odd, then \(x - y\) will still be even/odd. Similarly \((3y + 1)^5\) I reduce it to \(3y + 1\) and work with that instead. Now we can ask if \(x - y + 3y + 1 = x + 2y + 1\) is odd instead of asking if \((x - y)^3 + (3y + 1)^5\) is odd, because these two expressions must have the same parity (odd/even attribute).

Given x>y>0
S1.
x = 5(odd) & y =4(even)
(5-4)^3 + (3 x 4 +1)^5 = Even
x = 5 & y =3
(5-3)^3 + (3 x 3 + 1)^5 = Even

Confirmed NO

S2.
only Y is even but both options would change
if x is even/odd

Answer A