If $p$ and $q$ are odd, prove that there are no integral solutions Prove that if $p$ and $q$ are odd numbers , then the equation $x^{10} + p x^{​9} + q = 0$ does not have integral solution.
Could some hint a simple approach to solve this question. I am not getting how to initiate.
 A: Consider a parity argument.
If $x$ is even, then $x^{10}$ is even and $px^9$ is even, but $q$ is odd. So $x^{10} + px^9 + q$ is odd, but $0$ is even. So there are no solutions when $x$ is even.
If $x$ is odd, then $x^{10}$ is odd and $px^9$ is odd and $q$ is odd. But as the sum of three odd numbers is odd, we have again that $x^{10} + px^9 + q$ is odd, and so it cannot be zero.
This concludes the proof.

If you are familiar with modular arithmetic, looking mod 2 gives a slightly more straightforward and easier analysis. 
A: Hint: consider two cases: $x$ even and $x$ odd, and show that the left side is always odd.
A: One way utilizing some heavier machinery is to look at the equation $\mod 2 $. Observe then that since $p,q$ are odd that $p \equiv q \equiv 1 \mod 2$, so then it follows that we wish to solve 
$$ x^{10} + x^9 + 1 \equiv  0 \mod 2$$
But we already know that $$ x^k \equiv x \mod 2$$
So then it follows that we are solving
$$ x + x + 1 \equiv 0 \mod 2 \rightarrow 2x + 1 \equiv 0 \mod 2 \rightarrow 1 \equiv 0 \mod 2$$
And that is impossible, so we conclude the original equation has no solution.
Now I will re-write this in an algebraic way if you are not familiar with modular arithmetic
