Divisibility (2 Viewers)

seanieg89 · Apr 25, 2014

-y^2 can be 3, eg y=2.

HeroicPandas · Apr 25, 2014

seanieg89 said:
-y^2 can be 3, eg y=2.

sorry I made a mistake while finding out the quadratic residues (mod 7) (and another mistake for not checking it with yours and assuming that I'm right)

Shadowdude · Apr 25, 2014

let this be a lesson for me:

do not attempt maths in english mode.

my bad

HeroicPandas · Apr 26, 2014

Hello, 2 questions

1. What is that one-line method SpiralFlex?

2. seanig89, is my method now valid? I've fixed some mistakes, please look at my solution (red part)

obliviousninja · Apr 26, 2014

lol spiral is not srs

seanieg89 · Apr 26, 2014

HeroicPandas said:
Hello, 2 questions

1. What is that one-line method SpiralFlex?

2. seanig89, is my method now valid? I've fixed some mistakes, please look at my solution (red part)

Yep, all good. Your solution is the same as mine, just worded differently.

HeroicPandas · Apr 27, 2014

seanieg89 said:
Yep, all good. Your solution is the same as mine, just worded differently.

Yes I realized haha

Thanks for all the help!

anomalousdecay · Apr 27, 2014

HeroicPandas said:
No worries because I learn this in discrete maths (modular arithmetic)

Ah ok makes sense now because we did nothing like this in 1141.

SpiralFlex · Apr 27, 2014

My bad, I can't do it in one line.

$Q \rightarrow P: $As I said above, use that if$ \;a|b\; $and$ \;a|c\; $then for some integers$ \;s\; $and$ \;t,$

$we conclude$ \;a|sb+tc. \;$Prove of this is I don't think necessary but$ \;7|x \;$and$\; 7|y \;$implies$ \;7k=x, 7m = y. \;$Multiplying$\; (1)\; $by$\; s, (2) \;$by$\; t \;$and adding the equations we have,$

$7(k+m)=sx+yt, \;$so$\; 7|sx+yt.$

$\;$So if$\; 7|x \;$and$\; 7|y \;$then$\; 7|sx+ty. \;$Setting$\; s = x, t = y \;$gives$\; 7|x^2+y^2 \;$as desired.$$

$P \rightarrow Q:$ I thought I had a way involving using divisibility lemmas but it turned out to be wrong. So its probably best to use what Seanie posted. I will come back once I find a way without considering quadratic residues

HeroicPandas · Apr 27, 2014

SpiralFlex said:
My bad, I can't do it in one line.

$Q \rightarrow P: $As I said above, use that if$ \;a|b\; $and$ \;a|c\; $then for some integers$ \;s\; $and$ \;t,$

$we conclude$ \;a|sb+tc. \;$Prove of this is I don't think necessary but$ \;7|x \;$and$\; 7|y \;$implies$ \;7k=x, 7m = y. \;$Multiplying$\; (1)\; $by$\; s, (2) \;$by$\; t \;$and adding the equations we have,$

$7(k+m)=sx+yt, \;$so$\; 7|sx+yt.$

$\;$So if$\; 7|x \;$and$\; 7|y \;$then$\; 7|sx+ty. \;$Setting$\; s = x, t = y \;$gives$\; 7|x^2+y^2 \;$as desired.$$

$P \rightarrow Q:$ I thought I had a way involving using divisibility lemmas but it turned out to be wrong. So its probably best to use what Seanie posted. I will come back once I find a way without considering quadratic residues

q -> p, lol I forgot that, thanks for the revision

My method was basically the same thing:

$$ If $7|x $ and $7|y,$ then there exists positive integers A and B such that$ $ x = 7A and y = 7B. Squaring each and adding together gives \\ x^2 + y^2 = 49(A+B) = 7(M) $($where M = A+B \in \mathbb{Z}) \Rightarrow 7|x^2 +y^2$

Divisibility (2 Viewers)

seanieg89

Well-Known Member

HeroicPandas

Heroic!

Shadowdude

Cult of Personality

HeroicPandas

Heroic!

obliviousninja

(╯°□°）╯━︵ ┻━┻ - - - -

seanieg89

Well-Known Member

HeroicPandas

Heroic!

anomalousdecay

Premium Member

SpiralFlex

Well-Known Member

HeroicPandas

Heroic!

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