prime-power

**Sally_Math** · Jan 21st 2009, 08:36 AM

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.

**ThePerfectHacker** · Jan 21st 2009, 08:48 AM

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 $n>1$ )

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

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

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