Prove that the expression is multiple of $35$

Question

How to prove that the expression $15n^4 + 36n^2 + 106$ will be multiple of $35$ if $n$ is a natural number greater than $1$.

Answer 1

Not the case.

For simplification observe that:

$$15n^{4}+36n^{2}+106\equiv n^{2}+1\mod5$$

Then immediately we can conclude the expression is not a multiple of $5$ (hence not of $35$) if e.g. $n=4$.

Answer 2

It's not.

One big hint is that it doesn't work for $n = 1$. An integer polynomial which is divisible by some natural number $k$ for all $n$ over some bound is divisible by $k$ for all $n$.

For instance, we can immediately tell from the fact that $n = 1$ doesn't work that $n = 35+1$ doesn't work either: Expanding $15(35+1)^4 + 36(35+1)^2 + 106$ gives you a lot of terms that are divisible by $35$, and then what remains is $15+36+105$.

More concretely, for $n = 4$ we get $4522$ which isn't divisible by $5$, and therefore not by $35$.