# Proving an expression is not a perfect square

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!

• Hint: $a$ divides this product $\frac{a+b}{2}\frac{a+c}{2}\frac{b+c}{2}$ and so it divides at least one of the factors. Can $a$ divide either $\frac{a+b}{2}$ or $\frac{a+c}{2}$? – Stinking Bishop Aug 17 at 14:01
• Well, if it divides the first term then $\frac{a+b}2=ka$, which means that $a+b=2ka$, and $b=(2k-1)a$, which can only be true for $k=1$, otherwise $b$ wouldn't be prime. Interesting, let me see how I can use this. – Wolfuryo Aug 17 at 14:06
• Yes, and then with $k=1$ we would have $b=a$, while the problem states that $a,b,c$ are distinct primes. Now, do consider the third possibility, which is that $a$ divides $\frac{b+c}{2}$... – Stinking Bishop Aug 17 at 14:12
• I noticed that $k$ can't be 1, but I don't know how to proceed. If $a$ divides $\frac{b+c}2$, then it also divides $b+c$. Same thing can be said about $b$ and $a+c$, as well as $c$ and $a+b$. This means $a+c=kb$ etc., but I can't seem to be able to continue. – Wolfuryo Aug 17 at 14:22
• ... And, WLOG, presume $a$ is the biggest of the three prime numbers. How can it then divide $\frac{b+c}{2}$? – Stinking Bishop Aug 17 at 14:26

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 $$a>\frac{b+c}{2}$$.