If $t \in \mathbb{Z}$ then prove that $3t$ can never be a perfect cube except for $t=9$.
How can we prove things like these? I’m pretty new to Number-Theory and I find it difficult to prove things like these. Mathematical Induction cannot be used here because (beacuse the statement has an exception for $t=9$), even if we use a computer code then also it is not wise to iterate $t$ from $-\infty$ to $\infty$.
If we try to use proof by contradiction, our main question just gets transformed:
Let’s say there exists a $t$ other than 9 such that $$3t = n^3$$ Well this means $$t=\frac{n^3}{3}$$ that is there exists a perfect cube which is divisible by 3 (other than 27). But how to prove that there doesn’t exist any perfect cube which is divisible by 3 (other than 27).
I just need a hint for how to proceed for proofs like these.
Consider the number $t=3^2a^3$,
We have $$3(3^2a^3)=3^3a^3=(3a)^3$$
Clearly $a$ need not be $1$, hence there are many $t$ that make $3t$ a perfect cube.