Question

How to solve if
x^2 congruent to 7(mod11) has solutions?

I am sure there is something painfully obvious that I'm not seeing..

Answer 1

Yes, the tip of your nose.

Answer 2

if a ≡ b(mod m)

then by the def. of congruent modulo

m | (a - b)

then, if you let x = even integers

x = 2k

then, if you let x = odd integers

x = 2k + 1

then play around with these definitions, and the stars should align (hopefully)

Answer 3

a ≡ b(mod m)

is also equivalent to saying

a % m = b % m

but that should be obvious, and i dont think that definition will be as helpful.

Answer 4

@sarahusher Got it ?

Answer 5

i think maybe legendre is the way to go

Answer 6

@satellite73 Why not just standard definitions?

Answer 7

if \(\left(\frac{7}{11}\right)=-1\) then \(7\) is a quadratic residue if i recall correctly

Answer 8

No :( I'm still not getting it
in the legendre ^^ if it =-1 then it is a quadratic non residue

Answer 9

not sure how you go about finding the existence of square roots modulo a prime by standard definitions

Answer 10

oh right sorry you want 1 not -1

Answer 11

@satellite73 Well, I do; but lets have it your way.

Answer 12

kk i will listen

Answer 13

Oh okay, I get it in terms of the Legendre symbol now, thanks @satellite73

But I would also like to understand in the terms discussed above too, if that's okay? I want to make sure I fully understand it..

Answer 14

btw my calculation gives \(-1\)

Answer 15

that is \(7^5\equiv -1(\text{ mod }11)\)

Answer 16

and hello myininaya!!

Answer 17

Legendre, awesome. learn something new everyday.

Answer 18

Okay, yep I get -1 as well

Could someone explain to me the above method by using odd/even integers?

Answer 19

if we said 11 | (a - b) going off the defintiion

11 | (2k)^2 - 7 where we are letting x be even integers

11 | 4k^2 - 7

11 | 4(k^2 - 7/4)

11 | 4(k^2 - 7/4)

so the question becomes, does 11 divide (a-b) when a -b is a multiple of 4?

we find that if k = \[\frac{ 3 }{ \sqrt{2} } , \frac{ -3 }{ \sqrt{2} }\]

Answer 20

then, 11 | (a - b) is true, so perhaps that is sufficient to say x^2 congruent to 7 mod 11 has solutions

Answer 21

A similar approach would then be done with odds

Answer 22

@blurbendy No argument, @satellite73 gets the job done. You haven't seen my work yet, so you can't really compare can you?

Answer 23

touche monsieur!