The question is already mentioned in the title: For which odd number n does 503 divide $n^2+1$
$503\cdot q=n^2+1$
Asked
Active
Viewed 77 times
1 Answers
4
$503$ is a prime of the form $4k+3$ hence there is no integer $n$ such that $503$ divides $n^2+1$
Let $p=4k+3$ And assume that there is some integer $n$ such that $n^2\equiv -1 \mod p$ then we have $n^{p-1}\equiv (n^2)^{\frac{p-1}{2}}\equiv (-1)^{\frac{p-1}{2}}\equiv (-1)^{2k+1}\equiv -1 \mod p$
now using Fermat's little theorem we have $n^{p-1}\equiv 1 \mod p$ which is a contradiction.
Elaqqad
- 13,983
-
How can you conclude that? – Nick Podowalski Apr 11 '15 at 22:42
-
Do you know modular arithmetic? – Elaqqad Apr 11 '15 at 22:43
-
Yes, I do. Is it (mod 4)? – Nick Podowalski Apr 11 '15 at 22:44
-
1@NickPodowalski note that every square number is either $0$ or $1$ mod 4. So, $n^2+1$ is always either $1$ or $2$ mod 4. – JMoravitz Apr 11 '15 at 22:49
-
1@Igor you have made some mistake because it's not possible, see here – Elaqqad Apr 11 '15 at 23:04
-
Checked this in Excel :) All matches. Do you see my error? – Igor Deruga Apr 11 '15 at 23:10
-
@Igor The calculation in excel are not exact :$$94907047^2 + 1 =2+503\cdot 17907251630736$$ – Elaqqad Apr 11 '15 at 23:14
-
Yep, you're right. I guess, it's a double overfloat or something like that... Thanks! – Igor Deruga Apr 11 '15 at 23:18