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The question is pretty self-explanatory.I was wondering how this equation could be solved using "number theory".

Arian Tashakkor
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  • The equation transforms as $p^3-p^2=q^2+r^2=p^2(p-1)$, so we look for the set of sums of squares of two primes which are not square-free... – abiessu Apr 12 '15 at 16:25
  • @abiessu True.But I couldn't figure out the relationship between the primes. – Arian Tashakkor Apr 12 '15 at 16:28
  • I think we have the following:

    $$ p \mid p^3 \Rightarrow p \mid q^2, , , p \mid r^2 \Rightarrow p=q=r $$

    Furthermore, $p=q=r=3$ solves your equation.

     – Kevin Sheng Apr 12 '15 at 16:41
  • Are you expected to only use elementary methods or do you expect to use things like Gaussian integers here? – Bart Michels Apr 12 '15 at 16:46
  • @barto Elementary methods only – Arian Tashakkor Apr 12 '15 at 17:20
  • Your question reminds me of a question that I have previously asked here, though I am entirely sure that they are indeed related. I believe that at the beginning of this question, I have shown that with primes $[p,q,r]$ the only possible solution for $[p^3=p^2+r^2+q^2]$ is $[p=q=r=3]$, and with primes $[n,p,q,r]$ the only possible solution for $[n^4=n^2+p^2+r^2+q^2]$ is $[n=p=q=r=2]$. If it is indeed related as I suspect, then you might find it useful. – barak manos Apr 12 '15 at 20:37

2 Answers2

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Partial solution Note that $$p^3=p^2+q^2+r^2$$ implies that one of the numbers is equal to $3$, because if all of them are coprime with $3$ then $\mp1\equiv 1+1+1\mod 3$ which is absurd.

The case $p=3$ is easy,

The two remaining cases are symmetric , wa can assume that WLOG that $q=3$ then: $p^3-p^2=9+r^2$

Elaqqad
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  • The final equation is equivalent to $$(p-3)(p^2+p+3)=(p-r)(p+r)$$ or $$(p-3)(p^2+2p+6)=(r-3)(r+3)$$ if it helps someone – Elaqqad Apr 12 '15 at 17:03
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    IMO there's no need to post a partial answer only 12 minutes after the question is asked. If it remains unanswered for a day or so, then you can post a partial answer but surely at the time you wrote this you hadn't thought deeply about it yet. See meta. – Bart Michels Apr 13 '15 at 08:10
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The only deep result we need is the fact that primes congruent to $3$ mod $4$ remain primes as Gaussian integers.

To begin, it's easy to show that there are no solutions if $2$ is used as one (or more) of the primes; in particular, $8=4+q^2+r^2$ and $p^3=p^2+4+4$ have no solutions. So $p$, $q$, and $r$ are all odd. This implies

$$p\equiv p^3=p^2+q^2+r^2\equiv1+1+1=3\mod4$$

But writing the equation now as

$$p^3-p^2=q^2+r^2$$

implies $p\mid q\pm ir$ since $p$ is prime as a Gaussian integer, so we have $p\mid q$ and $p\mid r$, which is to say (since $q$ and $r$ are also primes), $p=q=r$. It quickly follows that $p^3=3p^2$ implies $p=q=r=3$ is the only solution.

Strictly speaking it's not necessary to invoke Gaussian integers; it suffices to know that $-1$ is not a quadratic residue mod $p$ if $p\equiv3$ mod $4$.

Barry Cipra
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