There exist no integers $a$ and $b$ for which $18a+6b=1$.
Proof: Assume that $18a+6b=1$. We find that
$$6(3a + b)=1$$
which leads to
We know that the sum of two integers can't produce a non-integer result, therefore a contradiction arises, as the proof demonstrates that two integers can produce a non-integer value. $\blacksquare$
My professor said that if one ends up with fractions in a proof, there is likely a problem. Can someone explain why this is the case?