# Re: Q's on Number Theory/Quadriatic Residues

```> >Bzzt!  Try Again.  If you use bc, you will notice that 9^2 mod 35 == 11
> >and 8^2 mod 35 == 29...  You should go take your number theory class!
>
> Definitely. Is there an easy way to get from the 29 to the 8?  I can see how
> it goes
> the other way, but what I didnt' see was how, if given 29, I could get the
> 8? (Euclid's?)

You can get the square root mod p (p prime) easily, if p+1 is divisible by 4.

You (should) know that  x ^ (p-1) equals to 1 mod p for every given x > 0.

Therefore  x ^ ( (p-1)/2 ) is either +1 or -1  mod p.

Now you have a given  x^2 and want to find x (one of both, there are two..)

( x^2 ) ^ ((p+1)/4)  =   x ^ (  (p+1)/2 )  =  x * x ^( (p-1)/2 ) = +/- x .

If p+1 is not divisible by 4, it's a little bit more difficult...

In your example, 35+1 is luckily divisible by 4.  But this doesn't help,
because 35 is not a prime.  35 = 5*7 , you can use the chinese remainder and
find the root mod 5 and the root mod 7. 7+1 is divisible by 4, you can use the trick:

( 8 ^ 2 ) ^ 2 mod 7 =  +/- 1.   (which is correct since 8 = 1 mod 7);

+1 and -1 are the roots of (8^2) modulo 7.

Modulo 5 we can't use the trick, but we guess the roots of (8^2) = 4 mod 5 as
2 and 3.

Back to the main problem: You want to have the root of (8^2) mod 35.

We found the roots  1 and 6 as roots of  (8^2) mod 7
and      the roots  2 and 3 as roots of  (8^2) mod 5.

Now solve the Chinese remainder for each possible pair (1,2), (1,3), (6,2), (6,3),
and you get the _four_ roots of (8^2) mod 35. Two of them will be 8 and -8, and there
are another two.

BTW: This is one way to do a mental coin flipping.