Basic algebra

[peak vigil]

The answer is six (I didnt get it)
I factorized stuff, thought of modulus and used odd/even logic.
3(x+y)^2 = (2+3x)(2+3y) - 4

Though I got it through graphing but I want to know the algebra way, that I am failing to find right now

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[timber tendon]

Off the top of my head I'd note that this is a quadratic equation.

You could use the quadratic formula to solve for one of the variables in terms of the other.

[short hill]

Maybe bringing it to the canonical form could help.

[wispy ridge]

Diophantine equation, is it?

[short hill]

If I'm not mistaken, the (close to) canonical form is ||(2x - y - 2)^2 + 3(y - 2)^2 = 16||, at which point we just need to test some cases.

[old spear]

goo goo gaa gaa

HIIIIIII

[old spear]

i have a few solutions

but we'll go with the straightforward one

[old spear]

Can you conclude the parities of x,y?

oh, even cooler

use the [unnamed] substitution

$\begin{cases}x=a+b\y=a-b\end{cases}$

[hushed fractal]

wolfqz

[short hill]

Well, that does kinda have a name.

[old spear]

ravi substitution or what

[short hill]

The name is "rotation and scaling" 😄

[old spear]

alright

[old spear]

obv you can go this way

or solve it as a quadratic in x

or let x,y be the solutions of the quadratic equation X^2-SX+P=0

these are the only 3 methods i can think of rn on no sleep

and all of them work

use whatever one you like

[short hill]

Or what I did above.

[old spear]

hmm similar to what happens when you do the rotation and scaling

[short hill]

Yeah, true. Though, technically, also translation.

[peak vigil]

Could you elaborate on the canonical form?

[short hill]

Basically, you complete squares until you're left with just constants.

If you're specifically working with second-degree curves/surfaces, you do substitutions after that, but that isn't necessary here.

Getting this form is enough.

[peak vigil]

Thanks

[short hill]

This isn't quite the canonical form, but it's a better form for this exercise (no fractions).

[peak vigil]

I see

[muted nacelle]

@peak vigil

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[muted nacelle]

@short hill @old spear @wispy ridge @timber tendon The user still needs help with this help request.

[old spear]

hi

[peak vigil]

Hi

I was unable to find the parity

[lapis knoll]

I know this may seem obscure, but what about denoting x and y to be roots of a quadratic, then using vieta’s formulae and newton’s sums concerting it into coefficients? Or maybe it’s far fetched?

[old spear]

that's what i suggested lol

[old spear]

.

[old spear]

just do casework.. if x and y are of opposite parity then x^2+y^2 is odd and 2(x+y)+xy is even

thus x,y are of the same parity

now check if they're both odd or both even

[lapis knoll]

I couldn’t view it

[old spear]

lmfao no issues

[timber tendon]

Like I said, just transform the equation into quadratic form to solve for one of the variables in terms of the other, then note that you require the discriminant to be a perfect square.

[old spear]

harder than just casework and bash smh