1. ## Equation manipulation

I was hoping that some of you with a lot more experience than me would check my work below for any blatant errors or incorrect assumptions, I have marked the parts I'm unsure about. I have deliberately included every step for clarity.

Two equations:

$b=u^2+4uv+2v^2$ ......... (1)

$a^2+c^2=2(u^2+2uv+2v^2)^2$............ (2)

Restrictions:

u and v are integers $u>v\sqrt{2}>0$
a, b and c are integers a > b > c > 0

I am trying to prove, given the 2 equations and restrictions, that there are no valid values for a, b or c. Proof that any one of them can't be an integer within the restrictions is enough.

From (2)

$\left(\dfrac{a+c}{2}\right)^2+\left(\dfrac{a-c}{2}\right)^2=(u^2+2uv+2v^2)^2$

manipulating the RHS (I'll call this Step 1 for a later question)

$\left(\dfrac{a+c}{2}\right)^2+\left(\dfrac{a-c}{2}\right)^2=(u^2+2v^2 )^2+4uv(u^2+uv+2v^2)$

$let\ \ \ \ \ \left(\dfrac{a+c}{2}\right)^2=(u^2+2v^2 )^2\ \ \ \ \ \ \ \ \ \ \ \ (3)$

$and\ \ \ \ \left(\dfrac{a-c}{2}\right)^2=4uv(u^2+uv+2v^2)\ \ \ \ \ \ \ (4)$

everything ok so far? (I think the min ratio of u:v has increased here, but not important at the minute)

from (3) .... $(a+c)^2=4(u^2+2v^2 )^2\ \ \ \ \ \ \ \ (5)$

$a+c=2(u^2+2v^2 )\ \ \ \ \ \ \ \ (6)$

from (4) .... $(a-c)^2=16uv(u^2+uv+2v^2)$

manipulating LHS

$(a+c)^2-4ac=16uv(u^2+uv+2v^2)\ \ \ \ \ \ \ (7)$

$(5)-(7)\ \ \ 4ac=4(u^4-4u^3v-8uv^3+4v^4)$

$ac=u^4-4u^3v-8uv^3+4v^4\ \ \ \ \ \ \ \ (8)$

from (6) .... $c=2(u^2+2v^2 )-a$

substituting this in (8) ...

$(2(u^2+2v^2 )-a)a=u^4-4u^3v-8uv^3+4v^4$

$a^2-2(u^2+2v^2 )a+(u^4-4u^3v-8uv^3+4v^4)=0$

$a=(u^2+2v^2 )+2\sqrt{u^3v+u^2v^2+2uv^3}$

for $a$ to be +ve integer $u^3v+u^2v^2+2uv^3$ must be a perfect square.

$let\ \ \ z^2=u^3v+u^2v^2+2uv^3$

no manipulation here, just grouping...

$z^2=(uv)^2+(u^2+2v^2)uv$

$(z+uv)(z-uv)=(u^2+2v^2)uv$

$so\ either\ z+uv=uv\ \ \ \ \ \ z=0$

$or\ \ z-uv=uv\ \ \ \ \ z=2uv$

are these two roots attained correctly?

$z=0\implies\ a=u^2+2v^2\implies\ (from\ (6))\ c=u^2+2v^2\ \ \therefore\ a=c\ so\ z\not=0$

$z=2uv\implies\ a=(u^2+2v^2 )+4uv\ \ \therefore\ a=b\ so\ z\not=2uv$

So this proves that $a$ cannot have a value that fits within the given constraints.

Question about Step 1. Does what I have shown only prove there's no valid $a$ when the equations are manipulated as in Step 1, or does it prove it conclusively?

If not conclusively, there are infinite ways of manipulating the equation at that stage $\left(i.e.\ \ \ (u^{100}+v^{250})^2+whatever\right)$

If this is the case, is there any way to prove conclusively what I'm attempting?

2. ## Re: Equation manipulation

$a+c = 2(u^2 + 2uv + 2v^2)^2$
divide both sides by 2 ...

$\frac{a+c}{2} = (u^2 + 2uv + 2v^2)^2$

... now, how did this next equation come about?

From (2)

$\left(\frac{a+c}{2}\right)^2 + \left(\frac{a-c}{2}\right)^2 = (u^2 + 2uv + 2v^2)^2$
are you saying $\frac{a+c}{2} = \left(\frac{a+c}{2}\right)^2 + \left(\frac{a-c}{2}\right)^2$ ?

3. ## Re: Equation manipulation

sigh, sorry, I checked and rechecked the op and missed that typo

I have edited the 2nd equation to the correct

$a^2+c^2=2(u^2+2uv+2v^2)^2$............ (2)

sorry again