Is a + or - if the sign of numbers $(-5)^{4n+2}a^{2n+5}b^{2n+1}c^{n+3}$ and $(-6)^{2n+3}a^nb^{2n-1}c^{n-5}$ is the same?

Question

Let a, b, c $\in \mathbb{Z}$ and $n\in \mathbb{N}$.

The question above is the the problem:

Is $\textbf{a}$ positive or negative integer if the sign of these numbers $(-5)^{4n+2}a^{2n+5}b^{2n+1}c^{n+3}$ and $(-6)^{2n+3}a^nb^{2n-1}c^{n-5}$ is the same?

$(-5)^{4n+2}$ will always be a positive number, because the exponent is always an even number. $(-6)^{2n+3}$ always negative, the exponent will be odd.

I tried to figure it out starting like this: 1. What happens if n is even? 2. And if n is odd?

Even if the exponents are odd or even, it doesn't guarantee the positiveness or negativeness of integer a.

Any tips?

Answer 1

Hint:

Exponents of $b$ and $c$ are both odd, does this cause a difference in the sign of both expressions. If it doesnt make a difference, then can you ignore them?

Answer 2

Consider the contribution each term makes either as $1$ or $-1$ depending on whether the value is positive or negative. If we can show evenness in the exponent, the term(or part of it) is positive no matter what.

We don't know what signs $a, b$ and $c$ have so whether the contribution is positive or negative we'll just use the unknowns themselves. Let's begin.

$(-5)^{4n+2}=(-5)^{2(2n+1)} \equiv 1$

$a^{2n+5}=a^{2(n+2)+1} \equiv a$

$b^{2n+1} \equiv b$

$c^{n+3}=c^{(n+1)+2} \equiv c^{n+1}$

$(-6)^{2n+3}=(-6)^{2(n+1)+1} \equiv -1$

$a^n \equiv a^n$

$b^{2n-1}=b^{2(n-1)+1} \equiv b$

$c^{n-5}=c^{(n+1)-6} \equiv c^{n+1}$

Remember that this equation is not of the numbers themselves but of their signs. It hence becomes:

$1 \cdot a \cdot b \cdot c^{n+1}=-1 \cdot a^n \cdot b \cdot c^{n+1}$

$b \cdot c^{n+1}$ appears on both sides and can be ignored. Our consideration then is for numbers $a$ such that:

$a=-a^n$

$n$ cannot be odd at a time like this because $a$ and $a^{n}$ have the same sign for all odd $n$. The negative sign destroys this possibility.

So we're at even values of $n$. $a^n$ then $has$ to be positive. $-a^n$ is hence negative.

You know what that means?

Do you?

$\boxed{a \text{ is negative!}}$