Problem 1
Let , where is an integer. Prove that cannot be expressed as the product of two non-constant polynomials with integer coefficients.
Step 4 of 4: Rule out the integer root
In plain words
An integer root would make the value zero, while parity says every integer value is odd.
Detailed analysis
Since the polynomial is monic and h is linear with integer coefficients, h would give an integer root of f. But for every integer x the displayed congruence makes f(x) odd, because the remaining constant term is 3. Therefore f has no integer root, a contradiction.