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We always have $\frac {p^2 + q^2}{pq} \geq 2$.

If $\frac {p^2 + q^2}{pq} = 2$ then this implies $(p-q)^2 = 0$, so as desired $p = q$.

I have not been able to make headway in the case $\frac {p^2 + q^2}{pq}= N$, where $N \geq 3$.

I suppose the argument required is to show that $p^2 + q^2 = Npq$ has no integer solutions in $p$ and $q$.

I tried considering $p$ and $q$ in $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/N\mathbb{Z}$ respectively, but failed to draw any further conclusions. Am I perhaps missing some key detail here?

I also tried supposing $p \neq q$ and saying without generality loss $p > q$. Then $p = q + k$ for some integer $k \geq 1$. This leaves us with $\frac {p^2 + q^2}{pq} = 2 + \frac {k^2}{pq}$. Now somehow I need to argue $k = 0$, and I am not quite sure how.

Or perhaps some other lateral approach is more apt, other than the ones suggested.

Aditya
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kapython
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4 Answers4

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Set $x=p/q$ and assume $n\in \mathbb{Z}_+$.

Then $\frac{p^2+q^2}{pq}=n$ implies $x$ is a (fully reduced!) rational solution to $$x^2-nx+1=0$$ From the rational roots theorem we know $p|1,q|1$ which means $p=1=q$.

Matthew H.
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$$p^2+q^2 = Npq$$ $$\implies$$ $$p^2 = (Np-q)q$$

So that $q$ divides $p^2$. Similarly, $p$ divides $q^2$. Now use that $gcd(p,q)=1$

Burno
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    I used Euclid's lemma directly on $p^2 = (Np - q)q$ and $q^2 = (Nq - p)p$ respectively to get what I sought. Out of curiosity: How do you argue after observing $p$ and $q$ divides $q^2$ and $p^2$ respectively, and that they are coprime? – kapython Jul 07 '23 at 14:11
  • I'm sure there are many ways. For example, since $p$ divides $q^2 = q.q$, then $\frac{p}{\text{gdc}(p,q)} = p$ divides $q$, and similarly $q$ divides $p$, so that they are equal (and since gcd($p$,$q$) $= 1$, $p = 1 = q$). – Burno Jul 07 '23 at 14:21
  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Jul 07 '23 at 16:23
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$\frac{p^2+q^2}{pq} =k \gt 2$

$\implies (p-q)^2=(k-2)pq$

Let $t$ be a prime dividing $p$. Then $t \ | \ (k-2)pq \implies t \ | \ (p-q)^2 \implies t \ | \ (p-q)$

But $t$ divides $p$ already, which means $t$ divides $q$ also, making $\text{gcd}(p,q) \gt 1$, contradiction.

Aditya
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The condition $\gcd(p,q)=1$ is not needed. If $m\in\Bbb N_0$ and $\sqrt m\in \Bbb Q$ then $\sqrt m\in \Bbb N_0$.

Now $\frac {p^2+q^2}{pq}=k\in\Bbb Z^+ \implies p^2-p(qk)+q^2=0 \implies p=(kq\pm q\sqrt {k^2-4})/2.$

This requires $\sqrt {k^2-4}\in\Bbb Q$ so $\sqrt {k^2-4}=n$ for some $n\in\Bbb N_0.$ This is not possible unless $n=0$ because there do not exist $k,n\in\Bbb Z^+$ with $k^2-n^2=4.$

Therefore $k=2$ and $0=\frac {p^2+q^2}{pq}-2=\frac {(p-q)^2}{pq}.$

A more elementary solution, assuming $\gcd(p,q)=1,$ is: If $r$ is a prime divisor 0f $p$ then

$r|p^2$ and $r|pq|p^2+q^2$

so $r|p^2+q^2$

so $r|(p^2+q^2)-p^2=q^2 $

so $r|q$ because $r$ is prime.

But then $\gcd(p,q)\ge r>1.$

So there is no such $r.$ So $p$ has no prime divisors. So $p=1.$

Interchanging $p,q,$ we also find $q=1.$

DanielWainfleet
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