is this a disproof?

P

procyon

Junior Member
Joined
Aug 13, 2011
Messages
53
I have a formula \(\displaystyle 8pq(p^2-q^2 )\)

I need to be able to find a \(\displaystyle u\) and \(\displaystyle v\) so that

\(\displaystyle 8pq(p^2-q^2 )=uv(4u^2-v^2 )\)

\(\displaystyle p, q, u,\) and \(\displaystyle v\) are all integers \(\displaystyle \not=0\)

I went about it like this

\(\displaystyle 8pq(p^2-q^2 )=8pq(p+q)(p-q)\)

\(\displaystyle uv(4u^2-v^2 )=uv(2u+v)(2u-v)\)

let \(\displaystyle 8pq=uv\).....(1)

and \(\displaystyle p+q=2u+v\)......(2)

and \(\displaystyle p-q=2u-v\).......(3)

(2)+(3) \(\displaystyle p=2u\).....(4)

(2)-(3) \(\displaystyle q=v\)......(5)

(4)x(5) \(\displaystyle pq=2uv\)

but we already have from (1) that \(\displaystyle 8pq=uv\)

Is this a proof that we can't convert \(\displaystyle 8pq(p^2-q^2 )\) into this format \(\displaystyle uv(4u^2-v^2 )\)

or is it just a proof that it can't be done this particular way?

Thanks for any advice
 
procyon said:
I have a formula \(\displaystyle 8pq(p^2-q^2 )\)

I need to be able to find a \(\displaystyle u\) and \(\displaystyle v\) so that

\(\displaystyle 8pq(p^2-q^2 )=uv(4u^2-v^2 )\)

\(\displaystyle p, q, u,\) and \(\displaystyle v\) are all integers \(\displaystyle \not=0\)

I went about it like this

\(\displaystyle 8pq(p^2-q^2 )=8pq(p+q)(p-q)\)

\(\displaystyle uv(4u^2-v^2 )=uv(2u+v)(2u-v)\)

let \(\displaystyle 8pq=uv\).....(1)

and \(\displaystyle p+q=2u+v\)......(2)

and \(\displaystyle p-q=2u-v\).......(3)

(2)+(3) \(\displaystyle p=2u\).....(4)

(2)-(3) \(\displaystyle q=v\)......(5)

(4)x(5) \(\displaystyle pq=2uv\)

but we already have from (1) that \(\displaystyle 8pq=uv\)

Is this a proof that we can't convert \(\displaystyle 8pq(p^2-q^2 )\) into this format \(\displaystyle uv(4u^2-v^2 )\)

or is it just a proof that it can't be done this particular way?

Thanks for any advice
Click to expand...

It can be done....

Try

u = p

v = 2q
 
procyon said:
I have a formula \(\displaystyle 8pq(p^2-q^2 )\)

I need to be able to find a \(\displaystyle u\) and \(\displaystyle v\) so that

\(\displaystyle 8pq(p^2-q^2 )=uv(4u^2-v^2 )\)

\(\displaystyle p, q, u,\) and \(\displaystyle v\) are all integers \(\displaystyle \not=0\)

I went about it like this

\(\displaystyle 8pq(p^2-q^2 )=8pq(p+q)(p-q)\)

\(\displaystyle uv(4u^2-v^2 )=uv(2u+v)(2u-v)\)

let \(\displaystyle 8pq=uv\).....(1) ..... Right here - unnecessary (and incorrect) constraint

and \(\displaystyle p+q=2u+v\)......(2)

and \(\displaystyle p-q=2u-v\).......(3)

(2)+(3) \(\displaystyle p=2u\).....(4)

(2)-(3) \(\displaystyle q=v\)......(5)

(4)x(5) \(\displaystyle pq=2uv\)

but we already have from (1) that \(\displaystyle 8pq=uv\)

Is this a proof that we can't convert \(\displaystyle 8pq(p^2-q^2 )\) into this format \(\displaystyle uv(4u^2-v^2 )\)

or is it just a proof that it can't be done this particular way?

Thanks for any advice
Click to expand...
.
 
Sorry, I don't follow :(

If (2) and (3) were correct then (1) has to be correct doesn't it?

This leads me to assume that (2) and (3) are also incorrect.

Although your answer is correct, I can't figure how you got it :(
 
also, would the method you used be able to do anything similar with this?

\(\displaystyle 4m^2-8mn-4n^2=4u^2+4uv-v^2\)

thanks again
 
procyon said:
Sorry, I don't follow :(

If (2) and (3) were correct then (1) has to be correct doesn't it?
Click to expand...

Who said (2) and (3) are correct?

It's not really a matter of correctness; it's more a matter of assumption.

You started by assuming that the relationships in (1), (2), and (3) were true, and then you proceeded based on those assumptions.

Your proof shows that the given information cannot be true based on those three assumptions.
 
Right, thank you.

I was just wondering if Subhotosh Khan had used a similar method or just excellent observation?
 
8pq(p^2-q^2 )=uv(4u^2-v^2 )​
8p^3*q - 8pq^3 = 4u^3*v - uv^3

Let

8p^3*q = 4u^3*v → v = 2q(p/u)^3

8pq^3 = uv^3

8pq^3 = u[2q(p/u)^3]^3

u^8 = [8q^3*p^9]/(8pq^3) → u = p & v = 2q
 
You must log in or register to reply here.
Top