@usukidoll A fiftieth power is a square.

@JasperLoy I have participated in English SE.

@MikeMiller Lee's Riemannian Manifolds. But it might be too elementary for you.

@Alizter Thanks, be sure to upvote me more often, lol.

@JasperLoy I already had one question however I have answered some stuff.

@Alizter Let me look.

@JasperLoy the your message in the starboard is weird :P

@UserX You're weird.

@UserX Thanks, I am weird.

@DanielFischer what are you talking about?

@JasperLoy where did you apply for gradschool?

we we're doing this in class and somehow the prof got $7^{32} = 92$ after we wrote 50 in base 2

how was that possible?

@UserX Not yet. I am trying to get well and study in 2015, and then take the GRE in 2016 and enter in 2017.

gre?

Is number theory taught in high-school/universities in the US?

universities

anyway just what is going on in this problem? I understand how it works, but that one spot is just grrrrrrrrrrrrrrrr!

It's a tie between number theory and analytic geometry in Greece and every teacher prefers analytic geometry because it can help you with complex numbers the next year

@UserX Well, some number theory is thought in elementary school, like prime factorisation, lol.

@JasperLoy Are my answers super bad?

@usukidoll Your question whether $7^{50}$ is a quadratic residue (modulo whatever). Since $7^{50} = (7^{25})^2$, the answer is yes.

then why was it that when we did it in class, the answer was no?

@usukidoll I wish I could help you but I don't know anything about number theory. Among the elementary/common fields, it's the field I haven't touched AT ALL along with combinatorics

@Alizter They seem pretty good, so 3 upvotes for you, lol.

@JasperLoy Boarder line serial upvoter :P

Any more and I will be accused of serial voting, lol.

@usukidoll Apparently, $7^{50}$ was computed (modulo $101$) to check whether $7$ is a quadratic residue modulo $101$ (which it is indeed not).

@Alizter You mean border.

@JasperLoy deletes English.SE account

@Alizter I am now trying to get 2000 on Eng SE as well, hopefully by Oct.

that is just... ok lost ... what we did in class was to figure out if $7^{50}$ was a quadratic residue of mod $101$ since $7^{50}$ was already a big number to deal with, the prof... broke it apart and then subtracted using multiples of 101 and then so many calculations later, the final answer was like 908 which was close to 909, but it was off by $-1$ so what is going on here and how did $7^{32}$ became $92$ @DanielFischer

@usukidoll No, you figured out whether $7$ was a quadratic residue modulo $101$ by computing $7^{50}$ modulo $101$. To check whether $7^{50}$ - or the remainder of that modulo $101$ - is a quadratic residue modulo $101$, you'd compute $(7^{50})^{50}$ modulo $101$ (unless you know the result beforehand and skip the computation).

EＨ？

@usukidoll You have the criterion that $a$ is a quadratic residue modulo the prime $p$ if and only if $a^{(p-1)/2} \equiv 1 \pmod{p}$.

YAH

sorry caps + I was typing in Japanese for some reason

Here, $p = 101$, so $(p-1)/2 = 50$.

Thus, to check whether $a$ is a quadratic residue modulo $p$, you compute $a^{50}$ modulo $p$.

hmm maybe another theorem was being used for that is $7^{50}$ a quadratic residue mod 101

or maybe no ?!

ugh

this is painful

oh yeah of course I would compute a^{50} mod $p$ but my a = 7 and p =101 and those are big fat numbers

If a number $s$ is a square - like any $a^{50}$ is - then the remainder of $s$ is a quadratic residue modulo all primes $q$ that don't divide $s$, since if $s = x^2$, then $x$ is a solution to the congruence $y^2 \equiv s \pmod{q}$.

@Chris'ssis ya I see that ... seems sos440 is on a coffee break :-)

I messed up on the title. The question is... is $7$ a Quadratic residue of mod $101$ obviously it's not, but I want to know how the heck is $7^{32} = 92?$ when the computation result was -9?

@usukidoll Because $92 \equiv -9 \pmod{101}$. $-9 + 101 = 92$.

how? ^

oh man -_-! so if we have a negative number, we add 101

?

You can do that, but need not. I would prefer to keep $-9$ here, since that has much smaller magnitude and is easier to calculate with.

I will eat something now, later.

@JasperLoy That sentence make no sense :P

who has played happy wheels?

@usukidoll In the past but I am out of that now.

@Alizter I am back now, lol.