# Is there any perfect square number can write in $p\cdot q\cdot r-p-q-r$? where $p,q,r$ are odd primes. [closed]

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])))))))

• See this post. We would have $p+q+r=8$ as a maximum, a contradiction to the assumption that $p,q,r$ are odd. Aug 7, 2021 at 16:22
• Thank you @DietrichBurde ,understand. Aug 7, 2021 at 16:34

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