3
$\begingroup$

Espress $2017$ as sum of two squares.

attempt: by Fermat's Theorem on sums of squares, the prime $p = 2017$ is the sum of two squares $2017 = a^2 + b^2$ , $a,b \in \mathbb{Z}$, if and only if $p \equiv 1 mod 4$.

And The irreducible elements in the Gaussian integers $\mathbb{Z[i]}$ are as follows $(a + bi)(a - bi) $ for primes $p\in \mathbb{Z}$ with $p \equiv 1 mod 4$ (both of which have norm $p$).

Then since $2017 \equiv 1 (mod 4)$ Then $2017 = a^2 + b^2$ .

Notice that $\sqrt2017 $ is approximately $44.91$. So $a^2, b^2 $ will be between values $1,2^2,....,44^2$ .

Then plugging different values from the above squares in $2017 - a^2 = b^2$
we find $2017 - 44^2 = 81 = 9^2$

So $2017 = 44^2 + 9^2$.

However, I found them using that approach. But is there a way to find them without doing this approach?

I dont' know how to use $p = a^2 + b^2 = (a + bi)(a - bi) $ for primes $p\in \mathbb{Z}$ with $p \equiv 1 mod 4$ (both of which have norm $p$).

So $2017 = a^2 + b^2 = (a+ bi)(a - bi) $. I don't' know how I would proceed assuming I would not have found the values . Any feedback or better approach would be appreciated it. Thank you!

$\endgroup$
1
$\begingroup$

I think I will throw in an advertisement for quadratic forms. Solve $u^2 \equiv -1 \pmod p.$ This could be by hand for small primes or primes of very special forms, otherwise it is Cornacchia or Tonelli-Shanks. Next we have $(2u)^2 \equiv -4 \pmod {4p},$ or $$ (2u)^2 = - 4 + 4 p t, $$ $$ (2u)^2 - 4 pt = -4. $$ This is a discriminant; we have the form $\langle p, 2u, t \rangle$ of discriminant $-4.$ The shorthand $\langle p, 2u, t \rangle$ means the (positive) binary quadratic form $$ f(x,y) = p x^2 + 2 u xy + t y^2. $$

Since this has discriminant $-4,$ it is equivalent by $SL_2 \mathbb Z$ to the only "reduced" form of that discriminant, namely $x^2 + y^2.$ In detail, given $$ G = \left( \begin{array}{cc} p & u \\ u & t \end{array} \right) $$ and $$ I = \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right) $$ there is a matrix $P$ of determinant $1$ such that $$ P^T G P = I. $$ Furthermore, it is very quick to find $P,$ this is usually called Gauss reduction. Next, take $$ Q = P^{-1}. $$ We then have $$ Q^T Q = G. $$ In particular $$ q_{11}^2 + q_{21}^2 = p. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.