# Is this proof for a and b being perfect squares correct?

Proof that if $k$ is an integer and $k^2=ab$ where a and b are natural numbers such that $HCF(a,b)=1$, a and b musty be perfect squares:

The $HCF(a,b)=1$ so a and b share no prime factors. Therefore $a$ and $b$ can be written as:

$a$=$p_1^{m_1} p_2^{m_2} p_3^{m_3} ... p_r^{m_r}$

and

$b$=(($q_1$)^($n_1$))(($q_2$)^($n_2$))(($q_3$)^($n_3$)...(($q_s$)^($n_s$))

where the different values $p$ and $q$ are primes and the different values of $m$ and $n$ are all integers.

Since $k^2=ab$,
$k^2$=(($p_1$)^($m_1$))(($p_2$)^($m_2$))...($p_r$)^($m_r$)*(($q_1$)^($n_1$))(($q_2$)^($n_2$))...(($q_s$)^($n_s$))

It has already been established that $a$ and $b$ share no prime factors so all of the values of $p$ an $q$ are unique. So

$k^2$=(($p_1$)^($m_1$))(($p_2$)^($m_2$))...($p_r$)^($m_r$)*(($q_1$)^($n_1$))(($q_2$)^($n_2$))...(($q_s$)^($n_s$))

is the prime factorisation of $k^2$. $k$ is an integer so $k^2$ is a perfect square. In the prime factorisation of a square number, the power of each prime is an even number so $m_1,m_2,m_3...m_r$ and $n_1,n_2,n_3...n_s$ are all even numbers.

Since

$a$=(($p_1$)^($m_1$))(($p_2$)^($m_2$))(($p_3$)^($m_3$))...($p_r$)^($m_r$)

and

$b$=(($q_1$)^($n_1$))(($q_2$)^($n_2$))(($q_3$)^($n_3$)...(($q_s$)^($n_s$)),

$a$ and $b$ must be perfect squares since the coeficient of each prime number in their prime factorisation is even.

• Seems correct to me. – ypercubeᵀᴹ Dec 30 '17 at 15:44
• This looks fine! – PJK Dec 30 '17 at 15:44
• Well explained. – Mohammad Riazi-Kermani Dec 30 '17 at 15:58