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I have the following proof:

Suppose it was a perfect square, then $\exists k$ such that $p^2 + pq = k^2$.

We can rewrite this as $pq = k^2 - p^2 = (k - p)(k + p)$. Since $p$ is a prime number, it must divide either $(k - p)$ or $(k + p)$. However, if $p$ divides $(k - p)$, then $(k + p)$ must be a multiple of $q$, which is not possible since $p$ and $q$ are distinct prime numbers. Similarly, if $p$ divides $(k + p)$, then $(k - p)$ must be a multiple of $q$, which is also not possible. Therefore, our assumption that $p^2 + pq$ is a perfect square leads to a contradiction, and we can conclude that $p^2 + pq$ is not a perfect square.

Is this proof correct? It feels a little weird and I don't know why

Anton Vrdoljak
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Jeff
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  • Why can't $k+p$ be a multiple of $q$? – TonyK Nov 05 '23 at 20:38
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    I might be wrong, but I don't understand why from $p^2 + pq = nk^2$ you ended up with an equation without the letter $n$: $pq = k^2 - p^2$. Where is the variable $n$? Maybe that is why it seems weird. – Daniel C. Nov 05 '23 at 20:38
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    The OP does not use $nk^2$ but rather $k^2$ for the rest of the answer. I think this is just a typo. – Sai Mehta Nov 05 '23 at 20:39
  • Ok, that is what I thought at first, thanks – Daniel C. Nov 05 '23 at 20:40
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    If $p^2+pq=k^2,$ then $k$ is divisible by $p.$ Hence $pq$ is divisible by $p^2,$ a contradiction. – Ryszard Szwarc Nov 05 '23 at 20:43
  • @RyszardSzwarc how do we go from $p \mid k$ to $p^2 \mid pq$? I see the contradiction but not the steps – Jeff Nov 05 '23 at 20:47
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    @Jeff: I suggest you spend a little time trying to work it out (it's not that difficult) instead of immediately asking for more help. – TonyK Nov 05 '23 at 20:48
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    @RyszardSzwarc: you also have to rule out the case $q=p$ (which would mean that $pq$ is divisible by $p^2$). – TonyK Nov 05 '23 at 20:50
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    @TonyK I took for granted that $p\neq q.$ If $p=q$ then $2p^2$ is not a perfect square as the multiplicity of the factor $2$ is either $1$ or $3.$ – Ryszard Szwarc Nov 05 '23 at 20:53

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This is not an answer to your proof-checking question, as I am not too fast at that. However, I would like to share with you this somewhat elegant proof.

Consider that:

$$ p^2 + pq = p(p + q) = k^2 \implies p \mid(p+q) \ \ \text{ in order to have a square} $$

But $pz = p + q \implies p \mid q$. Therefore $p = q$ necessarily, as distinct primes are coprime.

But if $p = q$ then in our original equation, we'd have:

$$ p^2 + p^2 = 2p^2 = k^2 $$

But $\Omega(2p^2) = 3$ while $\Omega(k^2) = 2\Omega(k) \in 2\Bbb{Z}$ (is an even number). $\blacksquare$

Where $\Omega : \Bbb{N} \to \Bbb{N}\cup 0$ is from number theory and counts (including multiplicities) the prime factors of a number. So $\Omega(9) = 2$, $\Omega(7) = 1$, $\Omega(24) = \Omega(2^3 \cdot 3) = 4$, etc. It is completely additive so $\Omega(k^2) = \Omega(k) + \Omega(k) = 2\Omega(k)$. A simpler way would be just to note that there always needs to be an even number of each prime dividing a square number.

Daniel Donnelly
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