High School Mathematics Extensions/Further Modular Arithmetic/Project/Finding the Square Root

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.. to be expanded There is actually a simple way to determine whether a is square

Let g be a generator of G where G is the multiplicative group mod p. Since all the squares form a group therefore, if a is a square, then

$a^{\frac{p-1}{2}}\equiv 1$

and if a is not a square then

$a^{\frac{p-1}{2}}\equiv -1$

we shall use these facts in the next section. .. to be expanded

*Finding the square root*

We aim to describe a way to find a square root in mod m. Let's start with the simplest case, where p is prime. In fact, for square root finding, the simplest case also happens to be the hardest.

If p ≡ 3 (mod 4) then it is easy to find a square root. Just note that if a has a square root then

$(a^{\frac{p+1}{4}}{)}^2\equiv a^{\frac{p+1}{2}}\equiv a^{\frac{p-1}{2}}a\equiv a$

So let us consider primes equivalent to 1 mod 4. Suppose we can find the square root of a mod p, and let

$x_0^2\equiv a(\mathrm{mod}p)$

With the above information, we can find the square of a mod p². We let

$x=x_0+x_{1}p$

we want x² ≡ a (mod p²), so

$x^2=x_0^2+2x_0{x}_{1}p+x_1^2{p}^2$

$x^2=a+kp+2x_0{x}_{1}p(\mathrm{mod}{p}^2)$

for some k as $x_0^2\equiv a(\mathrm{mod}p)$ so $x_0^2=a+kp$, we see that

$x^2=a+p(k+2x_0{x}_1)(\mathrm{mod}{p}^2)$

so if we need to find $x_1$ such that $k+2x_0{x}_1\equiv 0(\mathrm{mod}p)$, we simply need to make $x_1$ the subject

$x_1\equiv -k(2x_0{)}^{-1}(\mathrm{mod}p)$.

..generalisation ..example

..method for finding a sqr root mod p