Thread: prime-power

1) Prove that if n is a square, then each exponent in its prime-power decomposition is even.
and
2) Prove that if each exponent in the prime-power decomposition of n is even, then n is square.

Originally Posted by Sally_Math

1) Prove that if n is a square, then each exponent in its prime-power decomposition is even.
and

I do the first one and leave the second one for thee to think about.

(Assuming $\displaystyle n>1$)

If $\displaystyle n$ is a square it means $\displaystyle n=m^2$. Now $\displaystyle m>1$ and so it can be written as $\displaystyle m = p_1^{a_1}...p_k^{a_k}$ by prime decomposition. This means, $\displaystyle n = \left( p_1^{a_1}...p_k^{a_k} \right)^2 = p_1^{2a_1} ... p_k^{2a_k}$. And so exponents in prime decomposition of $\displaystyle n$ are even.

Similar problem: Once you prove the above problem try proving a stronger result. That $\displaystyle n>1$ is an $\displaystyle m$-th power if and only if each prime in the decomposition of $\displaystyle n$ is a multiple of $\displaystyle m$.