Question

@Kainui I will go ahead and pick a random problem from here so we can both just start somewhere and think about stuff. https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&cad=rja&uact=8&ved=0CDEQFjAC&url=http%3A%2F%2Fwww.math.muni.cz%2F~bulik%2Fvyuka%2Fpen-20070711.pdf&ei=a-pXVfuOEsm-sAWr5ICoBg&usg=AFQjCNESVFKertjKo1RRDenoVvOsgsul0w&sig2=vU0d9wa2LQ51WWsgH0873Q

Answer 1

Maybe number 5.

Answer 2

Awesome, I'll check it out and see if there's anything else interesting in there or if 5 gives me any ideas!

Answer 3

And yeah we are not just limited to 5 :p

Answer 4

There are a lot of tough and interesting ones here, just randomly looking around A10 and P1 sounds difficult, but I guess I should try to look into what "divisibility theory" even means.

Number 5 seems like a good place to start, I am playing with it a little but it's going to need some thought I think.

Answer 5

these are just thoughts
don't know if they will get me anywhere or not:
\[k xy=x^2+y^2+1 \\ \text{ we want to show } k=3 \\ 2xy+kxy=(x+y)^2+1 \\ \text{ or } kxy-2xy=(x-y)^2+1 \\ \text{ so we have } xy(2+k)=(x+y)^2+1 \text{ or } xy(k-2)=(x-y)^2+1\]

Answer 6

Oh that's a fun idea, I just tried something similar but it didn't work for me. All I did was replace \(x^2+y^2=(x+y)^2-2xy\) and stare at it blankly.

I'll play with what you've written since that seems like it might lead somewhere.

Answer 7

well that similar to what you just did

Answer 8

I added 2xy or -2xy on both sides

Answer 9

Here's another idea how about rewriting it as
\[\frac{1}{xy} + \frac{x}{y}+\frac{y}{x}=k\] don't know if that gets us anywhere either knowing that xy evenly divides I don' think we can say anything like each of these things in the sum is an integer unfortunately.

Answer 10

\[x^2+y^2=kxy-1 \]
maybe we can think about some Pythagorean triples

Answer 11

Oh how about considering this:

\[ \frac{x^2+y^2+1^2}{x*y*1}\]

Answer 12

\[3^2+4^2=5^2 \\ x=3,y=4, kxy-1=5^2 \\ 3\cdot 4 k-1=25 \\ 12k-1=25 \\ 12k=26 \\ k=\frac{26}{12}=3\]
so for (3,4,5) we have k=3

Answer 13

Oh that's interesting, isn't there some kind of formula for generating all pythagorean triples we could use to prove this generally for all x and y then?

Answer 14

oh I think it is called euclid's formula

Answer 15

\[(a,b,c)=(m^2-n^2,2mn,m^2+n^2)\]

Answer 16

ok so now do we just make a substitution into this formula and that's it? Hmmm

Answer 17

have to think more later
time is up

Answer 18

Haha alright.

Answer 19

I'll see ya soon then and think about this.

Answer 20

Since it seemed like an easy choice and potentially useful for completing the proof with Euclid's formula, if x=y then we have:

\[\frac{2x^2+1}{x^2} = 2+\frac{1}{x^2}\]

Since we are given that it evenly divides, then we know x=1 and the value is 3. Now what's still left to show is that if x>y the formula still holds true, which I'm sort of looking at connecting up with Euclid's formula that requires m>n in generating the triples once I figure it out.

Answer 21

26/12 =/= 3

Answer 22

that is a true statement dan but it doesnt really apply here hahaha

Answer 23

http://prntscr.com/764h9e

Answer 24

Oh yeah ok well that's a problem for that method I guess haha

Answer 25

hehe

Answer 26

How about this?

\(x^2+y^2+1 \equiv xy \mod 3\)
\(x^2+y^2-2 \equiv xy \mod 3\)
\((x+1)(x-1)+(y+1)(y-1) \equiv xy \mod 3\)

Hmmm I don't know this doesn't seem to lead anywhere.

Answer 27

so here's some stuff

\(0^2 \equiv 0 \mod 3 \)
\(1^2 \equiv 1 \mod 3\)
\(2^2 \equiv 1 \mod 3\)

so
\(x^2+y^2+1 \equiv 0 \mod 3\)
as long as x and y are not divisible by 3.

Answer 28

lol sometimes I fail at arithmetic

Answer 29

I don't know. I still think I need to play around with it more.
http://openstudy.com/users/ganeshie8#/updates/55470424e4b0868da1beb541
I think I might look at this a little.

Answer 30

oh you were actually on that question

Answer 31

fermat's infinite descent works like a charm for this problem

Answer 32

*

Answer 33

\[x^2-kyx+(y^2+1)=0 \\ \\ \text{ Let } a,b \text{ be roots } \\ a+b=ky \\ ab=y^2+1 \\ \text{ so we have } (a+b)^2=a^2+b^2+2ab=a^2+b^2+2(y^2+1) \\ (ky)^2=a^2+b^2+2(y^2+1) \\ k^2y^2=a^2+b^2+2(y^2+1) \\ k^2(ab-1)=a^2+b^2+2(y^2+1) \\ k^2ab=a^2+b^2+2(y^2+1)+k^2 \\ k^2=\frac{a^2+b^2+2(y^2+1)+k^2}{ab} \\ k^2=\frac{a^2+b^2+1}{ab}+\frac{2(y^2+1)+k^2-1}{ab} \\ k^2=k+\frac{2ab+k^2-1}{ab} \\ k^2=k+2+\frac{k^2-1}{ab} \\ k^2-k-2=\frac{k^2-1}{ab} \\ (k-2)(k+1)=\frac{(k-1)(k+1)}{ab} \\ k-2=\frac{k-1}{ab} \\ \\ k-2=\frac{k-2}{ab}+\frac{1}{ab} \\ 1=\frac{1}{ab}+\frac{1}{ab(k-2)} \\ ab=1+\frac{1}{k-2} \\ \text{ but } ab \text{ is an integer } \\ \text{ so } \frac{1}{k-2} \text{ is an integer } \\ \text{ either } k-2=1 \text{ or } k-2=-1 \\ \text{ so either } k=3 \text{ or } k=1 \\ \text{ so we have } ab=1+1=2 \text{ or } ab=1-1=0 \text{ but } a \text{ and } b \text{ are positive } \\ \text{ so } ab=0 \text{ is impossible } \\ \text{ so } k=3\]
still reviewing this
also some of the steps still may be wacky because I'm still trying to work out some thingys with it
I think ab cannot be zero
since x and y are positive
and f(x,y)=f(y,x)
where f(x,y)=(x^2+y^2+1)/(xy)

Answer 34

I should have never replace that one ab with y^2+1

Answer 35

when I said let a,b be roots
I mean where are function is f(x)=x^2-kyx+(y^2+1)
a quadratic in terms of the variable x
so f(a)=f(b)=0

Answer 36

even if we knew ab=0 was impossible
k still has to be positive
since
\[\frac{x^2+y^2+1}{xy}>0 \text{ since } x,y>0\]

Answer 37

even if we didn't know ab=0 was impossible*

Answer 38

looks neat! i couldn't find anything wrong...

Answer 39

still probably need to clean it up a little
and give a little more reason like why we can divide both sides by (k+1) on that one step
well we know k+1 isn't ever going to be zero since k is positive

Answer 40

\[x^2-kyx+(y^2+1)=0 \\ \\ \text{ Let } a,b \text{ be roots of the quadratic equation in terms of the variable } x \\I. ) a+b=ky \\ II. ) ab=y^2+1 \\ \text{ this next thing follows from squaring both sides of I.) } \\ (a+b)^2=a^2+b^2+2ab\\ \text{ Using } I.) \text{ again we have that we can replace } a+b \text{ with } ky \\ (ky)^2=a^2+b^2+2ab \\ k^2y^2=a^2+b^2+2ab \text{ using } II.) \text{ we can replace } y^2 \text{ with } ab-1 \\ \\ k^2(ab-1)=a^2+b^2+2ab \\ k^2ab=a^2+b^2+2ab+k^2 \\ k^2=\frac{a^2+b^2+2ab+k^2}{ab} \\ k^2=\frac{a^2+b^2+1}{ab}+\frac{2ab+k^2-1}{ab} \\ k^2=k+\frac{2ab+k^2-1}{ab} \\ k^2=k+2+\frac{k^2-1}{ab} \\ k^2-k-2=\frac{k^2-1}{ab} \\ (k-2)(k+1)=\frac{(k-1)(k+1)}{ab} \\ \text{ dividing both sides by } (k+1) \text{ since } k+1 \neq 0\\ k-2=\frac{k-1}{ab} \\ \\ k-2=\frac{k-2}{ab}+\frac{1}{ab} \\ 1=\frac{1}{ab}+\frac{1}{ab(k-2)} \\ ab=1+\frac{1}{k-2} \\ \text{ but } ab \text{ is an integer } \\ \text{ so } \frac{1}{k-2} \text{ is an integer } \\ \text{ either } k-2=1 \text{ or } k-2=-1 \\ \text{ so either } k=3 \text{ or } k=1 \\ \text{ so we have } ab=1+1=2 \text{ or } ab=1-1=0 \text{ but } a \text{ and } b \text{ are positive } \\ \text{ so } ab=0 \text{ is impossible since } ab=y^2+1>0 \text{ for all } y \\ \text{ so } k=3 \]

Answer 41

Ok I think this is a better and prettier version

Answer 42

still basically the same thing though

Answer 43

by the way about A10 I bet @rational would love that one
I have no clue how to do that one

Answer 44

I will make a new question for that one