number theory

**stumped765** · November 13th 2009, 04:06 PM

evaluate legender symbol

( 2 / 2^43112609 - 1)

ie ( 2/largest prime)

**Bruno J.** · November 13th 2009, 04:42 PM

There is no largest prime.

Use the fact that $(2/p)=1$ for $p\equiv\pm 1 \mod 8$ and $(2/p)=-1$ for $p\equiv\pm 3 \mod 8$ .

**ilikecandy** · November 13th 2009, 07:10 PM

Out of curiosity, how would you evaluate linear congruences like
$2^{43112609} - 1 \equiv a \mbox{ (mod 8)}$ if you have BIG numbers, without a calculator or computer?

**Bacterius** · November 13th 2009, 11:00 PM

If $a \equiv b$ (mod c), then $a^k \equiv b^k$ (mod c), I guess ?

**Bruno J.** · November 13th 2009, 11:09 PM

Originally Posted by ilikecandy

Out of curiosity, how would you evaluate linear congruences like
$2^{43112609} - 1 \equiv a \mbox{ (mod 8)}$ if you have BIG numbers, without a calculator or computer?

Think about it : $2^n \equiv 0 \mod 8$ for $n \geq 3$ . So $2^{43112609} - 1 \equiv -1 \mod 8$ .

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