Thread: is this a 'safe' assumption?

1. is this a 'safe' assumption?

Hi,

If I have the following equation

$8x^2-4y^2=z$

and it is known that x and y are both non-equal, non-zero, even integers

then z will even and can be written $z=2u$

so that $8x^2-4y^2=2u$

It is also required that z be a square integer

so it can be rewritten $z=2u=4v^2$

So with the above constraints I believe that

$8x^2-4y^2=z$

can be rewritten

$8x^2-4y^2=4v^2$

so

$2x^2-y^2=v^2$

Have I made any fundamental mistakes anywhere?

Many thanks

Pro

2. Re: is this a 'safe' assumption?

Given all the assumptions you made, that is correct.

3. Re: is this a 'safe' assumption?

Thanks HallsofIvy, glad to have someone check it over

Pro

4. Re: is this a 'safe' assumption?

What if y ≥ 2x ?

5. Re: is this a 'safe' assumption?

Originally Posted by SammyS
What if y ≥ 2x ?
Thanks SammyS.

That's true but the fact that z is a square implies $x>y/2$

6. Re: is this a 'safe' assumption?

Originally Posted by procyon
$8x^2-4y^2=z$
and it is known that x and y are both non-equal, non-zero, even integers
then z will even and can be written $z=2u$
so that $8x^2-4y^2=2u$
x and y can also be odd ... multiplications by evens (4,8 in your case)
always give even results....

7. Re: is this a 'safe' assumption?

Originally Posted by procyon
Thanks SammyS.

That's true but the fact that z is a square implies $x>y/2$
Usually the requirements are written at the beginning of the problem, so I missed the fact that z can be written as a square.

8. Re: is this a 'safe' assumption?

Originally Posted by Wilmer
x and y can also be odd ... multiplications by evens (4,8 in your case)
always give even results....
Thanks Wilmer.

That's true but x and y odd doesn't fit with other, simultaneous, equations I'm working with, so I thought I'd just state they're even in this case. x and y odd is valid though for a seperate set of equations.

Pro

9. Re: is this a 'safe' assumption?

Originally Posted by SammyS
Usually the requirements are written at the beginning of the problem, so I missed the fact that z can be written as a square.
Hi SammyS,

Yes, sorry about that. I was thinking whilst wrting instead of before

Pro

10. Re: is this a 'safe' assumption?

Originally Posted by procyon
That's true but x and y odd doesn't fit with other, simultaneous, equations I'm working with, so I thought
I'd just state they're even in this case. x and y odd is valid though for a seperate set of equations.
Disagree; x and y odd will work with the equations you've shown.
No separate set is required...let's see if Sammy agrees...

11. Re: is this a 'safe' assumption?

Originally Posted by Wilmer
Disagree; x and y odd will work with the equations you've shown.
No separate set is required...let's see if Sammy agrees...
I think if you re-read my previous reply, you'll find that I did actually agree with you on this

I'm actually saying that this equation was derived by setting other values to even

i.e.

$a=2b$ guarantees a is a positive even integer if b is any integer > 0
$a=2b+1$ guarantees a is positive odd integer if b is any integer ≥ 0

You might see the whole thing soon. At the minute I seem to have

2(odd integer) - (different [or same] odd integer) = even integer

which would prove what I'm trying to calculate is impossible.

damn, turned out to be odd, lol
[/edit]

Pro