I'm trying to solve this problem.
Is the number $x^9+2x^7+3x^3+4x$ (x belonging to $\Bbb Z$) divisible for 5? For which values of x, it is NOT divisible for 30?
My attempt. I tried to answer the first question by finding the equivalent $mod 5$ of each element.
- $x^9$ corresponds to x ($mod5$)
- $2x^7$ corresponds to $2x^3(mod5)$
So by thinking the expression mod5, then $x^9+2x^7+3x^3+4x = x+2x^3+3x^3+4x= 5x^3+5x$ which is reasonably a multiple of 5. Is my attempt correct? How can I answer the second question?
By FLT, $x^4\equiv 1\pmod 5$, so
$$x^9+2x^7+3x^3+4x\equiv x+2x^3+3x^3-x\equiv 5x^3\equiv 0\pmod 5$$
so yes, your approach is correct and the expression is divisible by $5$ for all $x\in\Bbb Z$
For divisibility by $30$, you need to check divisibility by $6$, ie, by $2$ and $3$. We have,
$$x^9+2x^7+3x^3+4x\equiv x+0+3x+0\equiv 4x\equiv 0\pmod 2$$
$$x^9+2x^7+3x^3+4x\equiv x+2x+0+4x\equiv 7x\equiv x\pmod 3$$
So, we have divisibility by $5$ and $2$ for all integers and divisibility by $3$ for those $x$ which are divisible by $3$. Hence, the expression is divisible by $30$ iff $x$ divisible by $3$