Congruence relation in radicals!

[Guest]

Hi dears .. i am newbie and register here a few seconds ago.

I've recently discovered something(I know it's just work on Integers, please don't animadvert)

[tex](\sqrt{5}-1)(\sqrt{5}+1)=4 \equiv -1(mod 5) \Leftrightarrow if \sqrt{5} \equiv 0 (mod 5) \Leftrightarrow (0-1)(0+1)=-1[/tex]

so we accept: [tex]\sqrt{5} \equiv 0 (mod 5)[/tex]

please tell me if you agree so as to i write here the rest.

and if you don't agree please criticize my deductive.

thank you so much. --{-@
(and sorry for my weak English)

[Guest]

Quadratic residues
http://en.wikipedia.org/wiki/Quadratic_residue
is the maths of studying equations of the form [tex]x^2 \equiv a \bmod n[/tex] (which is in some sense equivalent to trying to find [tex]x\equiv\sqrt{a}\bmod n[/tex]) this is probably what you want to look at.

You might be interested to know that not every integer has a "square root", for example [tex]sqrt{3}[/tex] doesn't work [tex]\bmod 8[/tex]. Also some integers have multiple roots, for example [tex]sqrt{1} \equiv 1[/tex], [tex]3[/tex], [tex]5[/tex], and [tex]7 \bmod 8[/tex].

Hope this helped,

R. Baber.

[Guest]

thx a lot for your answering.

furthermore, i think there are something else; Rational numbers! for ex. for [tex]\sqrt{3}[/tex] we can prove [tex]\sqrt{3} \equiv 0 (mod 3)[/tex]
or:
[tex](\sqrt{5}-1)(\sqrt{5}+1)=4 \equiv 0(mod 4) \Leftrightarrow if \sqrt{5} \equiv \pm 1 (mod 4) \Leftrightarrow (1-1)(1+1)=0 or (-1-1)(-1+1)=0[/tex]

finally we accept: [tex]\sqrt{5} \equiv \pm 1 (mod 4)[/tex]

that's not amazing?!