Finding pairs integers of a equation

[cerulean cypress]

What would be the best way to approach this?

[rapid vault]

Well, the equation is quadratic in both x and y, so what I'd do is use the quadratic formula to get one in terms of the other. Then the problem is reduced to "find integer values that yield integer values".

In particular, the discriminant must be a perfect square.

[cerulean cypress]

Like here? I just tried to do similar thing and couldn't find where

[rapid vault]

Well, do you want to find x or y?

[cerulean cypress]

Doesn't really matter as long I can get answer, but you can do it both ways? 😳

[rapid vault]

Yeah. It's either x^2 - (4y)x + (5y^2 + 2y - 4) = 0 or 5y^2 + (2 - 4x)y + (x^2 - 4) = 0.

[cerulean cypress]

Okay, thank you!

[cerulean cypress]

@rapid vault ah, the last question - after you solved and got an equation for x or y, is there a better way for finding them except just manually inputting numbers?

[rapid vault]

...I mean, did I not just describe a way better than guess and check?

[cerulean cypress]

Yes, but I have calculated -1 + 2x ± √[-(x-3)(x+7)]/5, but how I would proceed after that except just inputting numbers blindly knowing divisibility rules?

[rapid vault]

...I don't think you got that right.

[cerulean cypress]

uhh... Then how should I do it? I can replicate my calculations and send a screenshot here so it would be clearer what I did wrong

[rapid vault]

Yeah, that'd probably be a good idea.

[cerulean cypress]

@rapid vault here!

[rapid vault]

You corrected the thing you got wrong.

[rapid vault]

Remember what I said before, about the discriminant needing to be a perfect square?

Well. You corrected the thing you got wrong that I was talking about.

You also need parentheses on the numerator for the order of operations.

[cerulean cypress]

Yes, I double checked what I did and noticed that I got a bit lost in my notes 🥺

[rapid vault]

Anyway.

Discriminant. Perfect square.

[cerulean cypress]

But wouldn't it still be inputting manually to calculate the perfect squares?

[rapid vault]

Okay, look.

There's a difference between random guess and check and guided guess and check.

[cerulean cypress]

So it's still guess and check, but now I'm looking if numbers make perfect square, right?

[rapid vault]

Well. No.

Look.

Here's the thing.

Integers are real, right?

So what's the condition for y to be real here?

[cerulean cypress]

well, root should be non negative at the very least, if I understood your question correctly?

[rapid vault]

Right. Non-negative discriminant. So for what range of x is (3 - x)(7 + x) nonnegative?

We know the zeroes, so it's either the range between the zeroes or the range excluding the zeroes.

[cerulean cypress]

I feel like because of the minus sign before (3-x)(7+x) it should be negative (or zero)
So x is a range between [-7; 3]?

[rapid vault]

Well, you're right about the range. So x is an integer between -7 and 3. x is an element of the set {-7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3}. Which, while it is eleven elements, is still way fewer than the infinitely many elements we had before.

And we don't have to evaluate the whole value of y for each value of x.

And actually, this is symmetric.

[cerulean cypress]

Yes, I already made calculations with all of the numbers, however, in the end I got 7 pairs, but the answer for this question is 8. Did you, by any chance, solve it and if yes, what answer you got? I'm double-checking everything now

[rapid vault]

What pairs did you get?

[cerulean cypress]

(-7; -3), (-6; -2), (-5; -3), (-2; 6), (-2; 4), (1; 1), (3; -1)

Probably lost one somewhere in the middle of solving

[rapid vault]

You missed (2, 0).

Like I said, the values of x are symmetrical in what value they give the discriminant.

[cerulean cypress]

Ohh, yeah, I somehow forgot about 3-3

Thank you so much for helping and being patient! Sorry for being so slow!

[rapid vault]

At least you were paying attention and putting in effort.

[cerulean cypress]

+close