If$(10000x^{2015} - x)\over (x^2+x+1)$is odd, then $4x^2 + 3x + 1$ is even.

Question

How would I I begin to prove this implication? Starting with the hypothesis $(10000x^{2015} - x)\over(x^2+x+1)$ = $2(k)+1$.

I'm sort of lost on where to go?

Write x^{2015} instead of x^2015. Also use \frac{a}{b} for $\frac{a}{b}$. — user236182, Jan 12 '16 at 19:51
If $x$ is supposed to be an integer, we know that $x^2+x+1$ is odd (why?) and thus your assumption is just that $10000x^{2015}-x$ is odd, which means $x$ is odd, which means $3x+1$ is even. — Thomas Andrews, Jan 12 '16 at 19:59

score 2 · Answer 1 · answered Jan 12 '16 at 19:59

2

Suppose, $x$ is even. Then $1000x^{2015}-x$ is even and $x^2+x+1$ is odd. So, the fraction is even, contradicting the assumption.

Hence , $x$ must be odd and therefore $4x^2+3x+1$ even.

answered Jan 12 '16 at 19:59

Peter

score 1 · Answer 2 · answered Jan 12 '16 at 20:01

Look at the denominator first:

$$x^2+x+1=x(x+1)+1$$

At least one of $x$ and $x+1$ is even so $x(x+1)$ is even and $x(x+1)+1$ is thus odd .

This means that $$10000x^{2015}-x=(2k+1)(x^2+x+1)$$ is odd because it's the product of two odd numbers .

But $10000x^{2015}$ is even so $x$ must be odd .

To finnish notice that $4x^2$ is even and $3x$ and $1$ are odd . Thus the number $$4x^2+3x+1$$ is even .

2 Answers2