Solve explicitly integer solution

[annoymage]

Homework Statement

find p,q,r,s integer solution

$$\frac{1}{p^2}+\frac{1}{q^2}+\frac{1}{r^2}+\frac{1}{s^2}=1$$

Homework Equations

here some alternate form you can see

http://www.wolframalpha.com/input/?i=1/p^2%2B1/q^2%2B1/r^2%2B1/s^2%3D1

The Attempt at a Solution

i don't know if this works,

so i guess i have to show that $$p=q=r=s$$,

so now i only got this $$p|(qrs)^2\ \ ,\ q|(prs)^2\ \ ,\ r|(pqs)^2\ \ ,\ s|(pqr)^2$$, but i don't even know how to show $$p|q$$

help T_T

 

[Office_Shredder]

You should just do a size comparison

One side is: Can any of p, q r or s be equal to 1?

And then the flip side: What happens if p is a large integer (how are the possible sizes for q, r and s restricted)

 

[annoymage]

One side is: Can any of p, q r or s be equal to 1?

it can't, i'll show to you to check my proof of this later

And then the flip side: What happens if p is a large integer (how are the possible sizes for q, r and s restricted)

hmm, do you mean p>2??

if yes, i only get $$\frac{1}{q^2}+\frac{1}{r^2}+\frac{1}{s^2} < \frac{1}{2}$$ how to get restriction on p,q,r?

 

[Office_Shredder]

If p>2, you should get the inequality $$\frac{1}{q^2}+\frac{1}{r^2}+\frac{1}{s^2}>\frac{3}{4}$$

 

[annoymage]

yeaa that's true, maybe i don't understand the flip flip side thing. very sleepy, i'll try understand it tomorrow. anyway, what should i do next?

assume some more what happen if q>2 ??

 

[Office_Shredder]

Well, if none of q,r and s are 1, what's the largest $$\frac{1}{q^2} + \frac{1}{r^2}+\frac{1}{s^2}$$ can be?

 

[annoymage]

sorry i still don't get it

hmm, i guess there's nothing to do with post 3 and 4,

this is what i understand p,q,r,s can't be 1

if p>1 then largest $$\frac{1}{q^2} + \frac{1}{r^2}+\frac{1}{s^2}$$ is 3/4

then? ;P

 

[Office_Shredder]

Use what's in post number 4 (which you should work on solving for: why don't you post your attempt at it?)

 

[annoymage]

aahhh, maybe i see now,

so p=q=r=s=2 is a solution.

so that we goin through is showing that p not equal 2 in Z+ is not a solution.

also i have to repeat for q,r and s right?

 

[Office_Shredder]

If there is a solution for which q (or r or s) is not equal to 2, you should be able to prove in a line or two that a solution exists for which p is not equal to 2 (which gives a contradiction of course)