iff proof

Hello,

Write the decomposition into prime factors of n.

$n = \prod_{i=1}^k p_i ^{\alpha_i}$

So $n^2 = \prod_{i=1}^k p_i ^{\alpha_i} \prod_{i=1}^k p_i^{\alpha_i} = \prod_{i=1}^k [p_i^{\alpha_i} \times p_i^{\alpha_i}]$

As $a^b a^c = a^{b+c}$ , the previous expression equals to :

$\prod_{i=1}^k p_i^{\alpha_i + \alpha_i} = \prod_{i=1}^k p_i^{2 \alpha_i}$

Hence the exponent of every prime number occuring in the factorisation of n is even because multiple of 2.

And as it's equalities, there is equivalence.