Is There a Solution for x^4 ≡ 2 (mod p) When p ≡ 1 (mod 4)?

[b0mb0nika]

show that x^4 == 2( mod p) has a solution for p==1(mod 4) iff
p is of the form A^2+64B^2, where A,B are integers


I let x^2=M
then the conguence is reduced to M^2==2( mod p)
but any # squared == 0 or 1 ( mod4 ) so p must be == 1(mod4)...
but I'm not sure what to do now..
any hints/ or ideas ?

 

[matt grime]

Find when 2 is a residue mod p using, say, Quadratic Reciprocity, and then find when its square root is also residue, again using reciprocity

 

[M98Ranger]

To solve this problem, we need to use the properties of biquadratic residues and the fact that p is congruent to 1 modulo 4.

First, let's recall that a biquadratic residue mod p is a number that can be expressed as the square of another residue mod p. In other words, if x^2 is a residue mod p, then x^4 is also a residue mod p.

Now, let's consider the congruence x^4 ≡ 2 (mod p). Since p is congruent to 1 modulo 4, we know that p ≡ 1 (mod 4). This means that p can be expressed as p = 4k + 1 for some integer k.

Substituting this into our congruence, we get x^4 ≡ 2 (mod 4k + 1). We can then use the property of biquadratic residues to rewrite this as (x^2)^2 ≡ 2 (mod 4k + 1).

Now, we can use the fact that 4k + 1 is congruent to 1 modulo 4 to further simplify the congruence to (x^2)^2 ≡ 2 (mod 4).

We know that for any integer n, n^2 ≡ 0 or 1 (mod 4). So for this congruence to have a solution, we need 2 to be congruent to one of these values mod 4. Since 2 is not congruent to 0 modulo 4, it must be congruent to 1.

This means that (x^2)^2 ≡ 1 (mod 4), which implies that x^2 ≡ ±1 (mod 4).

Now, we can use the fact that p is congruent to 1 modulo 4 to simplify this even further. Since p = 4k + 1, we can rewrite x^2 ≡ ±1 (mod 4) as x^2 ≡ ±1 (mod p).

This means that x^2 is a biquadratic residue mod p. And since p is congruent to 1 modulo 4, we know that p can be expressed as p = A^2 + 64B^2 for some integers A and B.

So, we have shown that if p is