0
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ 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 $\frac{a+c}{2}$? $\endgroup$ – Stinking Bishop Aug 17 at 14:01
  • $\begingroup$ 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. $\endgroup$ – Wolfuryo Aug 17 at 14:06
  • 1
    $\begingroup$ 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 $\frac{b+c}{2}$... $\endgroup$ – Stinking Bishop Aug 17 at 14:12
  • $\begingroup$ 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. $\endgroup$ – Wolfuryo Aug 17 at 14:22
  • 2
    $\begingroup$ ... And, WLOG, presume $a$ is the biggest of the three prime numbers. How can it then divide $\frac{b+c}{2}$? $\endgroup$ – Stinking Bishop Aug 17 at 14:26
2
$\begingroup$

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 $a>\frac{b+c}{2}$.
| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.