26. This exercise and Exercises 27 and 28 give Dirichlet's beautiful proof that 2 is a biquadrati...

Question

Question

Answer 1

Answer #1

(a) Since $p=a^2+b^2$ with $a$ odd, we have

$\begin{pmatrix}{\frac ap}\end{pmatrix}=(-1)^{{\frac{p-1}2}{\frac{a-1}2}}\begin{pmatrix}{\frac pa}\end{pmatrix}$

by quadratic reciprocity of Jacobi. Since $p, d$ odd and $p1 mod 4$ , we know that

$p-1$

is even; therefore,

$(-1)デデ$

Also, $p=a^2+b^2$ implies

$(:) = 1$

Therefore,

$(5) = (-1)デデ/L)-1$

(c) From $p=a^2+b^2$ we get

$p-a2 + b2-(a + b)2-2ab (a + b)2-2ab mod p 200$

(d) From part c) we get

$2a6 mod p a$

Hence,

$(a + b)宁ー(21)$

Answer 2