10 Replies
@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.@coolguy. mod malum hai?
Imma elaborate on the top answer that was given on the site:
$4n^{2}+1 \equiv 0 (mod 5)$ which means since $4 \equiv -1 (mod 5)$, we get $(-1)n^{2}+1 \equiv 0 (mod 5) \implies 1 \equiv n^{2} (mod 5)$. Similarly for 13, $4n^{2}+1 \equiv 0 (mod 13)$ which means $4n^{2} \equiv -1 (mod 13)$ which means $n^{2} \equiv 12/4=3 (mod 13)$
SirLancelotDuLac
For $n^{2}$ to be 3 mod 13, n would have to be congruent to + or -4. (Because 3 is congruent to 16 and $n^{2} \equiv 16$ means $n \equiv 4/(-4)$)
SirLancelotDuLac
After that use chinese remainder theorem and a bit of case-working to get the desired n which satisfy the condition, which are infinite in number.
Ahh kayyy
Yeah yeah
Ima try and get back
+solved @SirLancelotDuLac
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