Number Theory | When is there a square root of 2 modulo p?

We classify the primes for which 2 is a quadratic residue.

[Sam-gfir]

Wow! this is the coolest math channel ever!

[udichonstkovski]

1:45 you are saying k is between 0 and p-1. you are writing k is greater or equal to 1 which is correct, but k should also be smaller than (p-1)/2 and that makes the rest more sensible.
(if k<p-1, then for example let p=19, then 2k (mod 19) < 19/2 = 9.5 where 10=<k<15 is also valid . since 20 mod 19 is 1, 22 mod 19 is 3 etc. )

[Blabla]

Hold on, something goes wring here.

1. typo at 1:42: p-1 should be (p-1)/2.
2. at 2:59: 2k < p/2 <=> k < p/4. This is incorrect since that's not what it originally said. It should say: 2k mod (p) < p/2 and from this it does not follow k < p/2

A fix to this is: if k < (p-1)/2 then 2k < p-1 and thus 2k (mod p) = 2k (the mod doesn't do anything) and the original argument (2) can continu.

[georgesadler]

Professor M. Penn, thank you for an excellent Number Theory video.

[walidabdelal]

Nice, Michael! Also, Apostol’s Introduciton to Analytic Number Theory has a nice proof of the general form in Theorem 9.5 of Ch.9 §9.3

[demenion]

so -1 is a QR iff p²=1 (mod 4) and 2 is a QR iff p²=1 (mod 8). Are there similar special cases for p²=1 (mod 2^k) for k>3?

[harshshah]

I don't know what should i say but somehow there are strings attached betwn us I want to communicate with u