TheSimpliFire's Chatroom

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8:43 AM

@Peter

checked up to $10^9$:

for(n=1, 10^9,   a= 1-4*n; t= 1;   forprime(p=2, n,    if(kronecker(a,p)==-1, t= 0; break());   );   if(t, b= 4*n!+a; if(issquare(b), print(n"    "a"    "b)));  )

— Dmitry Ezhov 7 hours ago

1 hour later…

Peter

9:47 AM

@TheSimpliFire I have doubts about this code, both because of the time needed to calculate the factorials (the number n! is calculated directly each time which is not efficient) and the size of the numbers involved. I had memory problems with my pari/gp version and maximum chosen allocation with much smaller numbers than $(10^9)!$. It might be that my computer is too old , but I have difficulties to take the claim serious that he actually checked the claim upto $n=10^9$

TheSimpliFire

@Peter I don't get what if(t, b= 4*n!+a; means

t isn't a condition

Peter

10:05 AM

yes, this is weird as well.

Could you perhaps continue the factor search ?

TheSimpliFire

In a few hours

Peter

Haran

10:39 AM

A: Can $4\cdot n!-4n+1$ be a perfect square when $n>4$?

If $\nu_p(4n-1)< \nu_p(n!)$ for all primes $p \mid (4n-1)$, then we have- $$x^2 = 4n!-4n+1=(4n-1)(\frac{4n!}{4n-1}-1)$$ and for every $p \mid (4n-1)$, we also have $p \mid \frac{4n!}{4n-1}$. Thus- $$\gcd\bigg(4n-1,\frac{4n!}{4n-1}-1\bigg)=1 \implies (4n-1)=x_1^2$$ However, this is impossible as $...