Question

If $c^2=ab$and $\left(a,b\right)=1$, prove that a and b are perfect squares.

Short answer

It is proved that a and b are perfect squares.

$a={\left(\prod _{i=1}^k{p_i}^{\frac{r_i}{2}}\right)}^2$

$b={\left(\prod _{j=1}^l{p_j}^{{\displaystyle \frac{s_j}{2}}}\right)}^2$

Step-by-step solution

Define a and  b

Consider, $a=\left(\prod _{i=1}^k{p_i}^{r_i}\right)$and role="math" localid="1646217449793" $b=\left(\prod _{j=1}^l{p_j}^{s_j}\right)$

Here, a and b are the prime numbers, and these numbers are the unique decomposition of a and b.

If $r_i,s_j\ge 1$, then condition $\left(a,b\right)=1$indicates that they are distinct from the prime numbers $p_i$and $q_j$ within the decomposition of a and b , respectively.

Obtain c2=ab

Consider the given equation $c^2=ab$; if $c|ab$, then c can be written as

role="math" localid="1646217958566" $c=\left(\prod _{i=1}^k{p_i}^{r_i}\right)\left(\prod _{j=1}^l{p_j}^{s_j}\right)$.

Further solving,

$c^2=ab$

The above equation is equivalent to $2r_i=r_i$, and $2s_j=s_j$. Here $r_i$ and $s_j$are the decomposition of a and b , and all are even.