-2
$\begingroup$

Following is an experimental claim

Can it be shown that

Let $p, q$ and $r$ are odd primes then there is no perfect square number $n$ can express in the form as $p\cdot q\cdot r-p-q-r$?

Example: $100$ can't express for any prime $p, q,r$

I have checked up to 30000, Source code PARI/GP

for(u=1,30000,forprime(p=3,10,forprime(q=p,30,forprime(r=q, 100,if(p*q*r-p-q-r==u,if(issquare(u)==1,print([u,p,q, r])))))))
$\endgroup$
2
  • 1
    $\begingroup$ See this post. We would have $p+q+r=8$ as a maximum, a contradiction to the assumption that $p,q,r$ are odd. $\endgroup$ Aug 7, 2021 at 16:22
  • $\begingroup$ Thank you @DietrichBurde ,understand. $\endgroup$
    – Pruthviraj
    Aug 7, 2021 at 16:34

1 Answer 1

3
$\begingroup$

If $p,q,r$ are odd primes each is $\equiv \pm 1 \bmod 4$. Hence the product $pqr \equiv \pm 1 \bmod 4$.

Case 1: $pqr \equiv 1 \bmod 4$. Then either $p\equiv q \equiv r \equiv 1 \bmod 4$ or WLOG $p\equiv 1 \bmod 4$ and $q \equiv r \equiv -1 \bmod 4$ (other permutations do not alter the logic). Thus, either $$pqr-p-q-r \equiv 1-1-1-1=-2 \bmod 4$$ or $$pqr-p-q-r \equiv 1-1-(-1)-(-1)=2 \bmod 4$$ In either case $n=pqr-p-q-r \equiv \pm 2 \bmod 4$ and it is known that square numbers $\equiv 0,1 \bmod 4$, so in Case 1, it is impossible for $n$ to be a square.

Case 2: $pqr \equiv -1 \bmod 4$. Then either $p\equiv q \equiv r \equiv -1 \bmod 4$ or WLOG $p\equiv -1 \bmod 4$ and $q \equiv r \equiv 1 \bmod 4$ (other permutations do not alter the logic). Thus, either $$pqr-p-q-r \equiv -1-(-1)-(-1)-(-1)=2 \bmod 4$$ or $$pqr-p-q-r \equiv -1-(-1)-1-1=-2 \bmod 4$$ For the same reason as in Case 1, it is impossible that $n=pqr-p-q-r$ is a square.

Note that the primeness of $p,q,r$ is unnecessary to the result

$\endgroup$

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