Proving an expression is not a perfect square

Question

Let a, b and c be 3 odd, distinct prime numbers. I have to prove that the product $abc\frac{a+b}2\frac{a+c}2\frac{b+c}2$ cannot be a perfect square.

Since a, b and c are prime, we have that $\frac{a+b}2\frac{a+c}2\frac{b+c}2=k^2abc$, with $k$ natural for the product to be a perfect square. I tried applying AM-GM on $\frac{a+b}2\frac{a+c}2\frac{b+c}2$ and $abc$ but it didn't get me anywhere.

I'm pretty sure this is supposed to be solved with pretty basic theory, but I'm not sure. Thanks for your help!

Answer 1

Suppose $K=abc\frac{a+b}{2}\frac{a+c}{2}\frac{b+c}{2}$ is a perfect square, with $a, b, c$ - distinct odd primes.

Without loss of generality, let $a$ be the biggest of the three primes.

As $a$ divides $K$, and $K$ is a perfect square, then $a^2$ must also divide $K$, so $a$ must divide $bc\frac{a+b}{2}\frac{a+c}{2}\frac{b+c}{2}$. Being coprime to $b$ and $c$, we conclude that $a$ divides one of: $\frac{a+b}{2}$, $\frac{a+c}{2}$, $\frac{b+c}{2}$. However, this is impossible, as:

• $a$ cannot divide $\frac{a+b}{2}$ because otherwise $a$ would also divide $a+b$ and therefore $a$ would divide $b$.
• Similarly, $a$ cannot divide $\frac{a+c}{2}$.
• Finally $a$ cannot divide $\frac{b+c}{2}$ either, because it is too big. As we assumed $a$ to be the biggest, i.e. $a>b, a>c$, we also have $a>\frac{b+c}{2}$.