if x = u^2 - v^2, y = 2uv, and z = u^2 + v2, and if x = 11

[NL]

both of those pairs do, in fact, satisfy both equations (and, moreover, they also satisfy u^2 - v^2 = 11).

i'm not sure where you are going from there.

No, Ron, I’m still alive, vividly!
Yes, I forgot the goal, which is u^2+v^2. So yes, negative or positive numbers don’t matter. You’re… great.

But I have a doubt: How do we know that only (5, 6) and (-5, -6) satisfy the 2 equations? Can’t we get some fractions? (I know it’s little dumb, but I have no knowledge to confirm this doubt)
- I try to answer: Let’s get simpler numbers, such as:
(1) 3*1 =3 and
(2) 3^2-1^2 = 8

- Can we get two numbers that satisfy (1)? Sure:
1/3 * 9 or 1/9* 27

- Why those numbers can’t satisfy (2)?
If we square both numbers, 9^2 will go far away from 3^2 to the right (of number 1) and (1/3)^2 will go far away from 1/3 to the left (of 1), so the distance between 9^2 and (1/3)^2 will surely be bigger than 8. So we can’t get a distance equal to 3^2-1^2.

For other fractions, distances will be bigger because for a bigger denominator, we need a bigger numerator to offset.

It’s also true if the (-) equation is replaced with + or / (maybe :))

Any thing wrong in my reasoning?
Any special number that doesn’t reflect my conclusion drawn from the testing case above?

p/s: sorry for a lot of grammatical errors. Just fixed some :)

[RonPurewal]

ok, well, first of all, you wrote this:

I’m likely to stop at this point:
uv =30
u^2-v^2 =11

It seems like either (u,v)= (6,5) or (-6, -5) is satisfying both equations. Any way to know that one pair is not satisfying without actually solving the “giant” equation?

the pink stuff makes me think you're saying "i can stop here because i'm thinking 'not sufficient'."
if that's the case, then, no you can't "stop at this point", since both of these results give the same value for z (= the actual goal of the problem).

[RonPurewal]

in any case, you can solve statement 1 with numerical intuition, if you have that sort of intuition.

here's what i mean:

• you have u^2 - v^2 = 11.

• this is a constant difference of 11. therefore, if u^2 gets bigger, then v^2 will have to get bigger along with it, as an offset to keep the difference = 11.

• as u^2 and v^2 get bigger, the magnitude (absolute value) of 2uv will get bigger, too.

• so, there is going to be a unique pair of u^2 and v^2 that can satisfy u^2 - v^2 = 11 and also 2uv = 60.

this is a nice lazy person's solution, but, like most other "lazy" solutions, it requires a rather deep intuitive sense.

[NL]

• you have u^2 - v^2 = 11.

• this is a constant difference of 11. therefore, if u^2 gets bigger, then v^2 will have to get bigger along with it, as an offset to keep the difference = 11.

Oh, that’s superior. I didn’t see this. We even don’t have to test numbers. I think “Thursday with Ron” should have a slogan: “Discover countless ways of thinking with Ron”. Cool!

• as u^2 and v^2 get bigger, the magnitude (absolute value) of 2uv will get bigger, too.

Yes, like if 2 legs grow longer, the pan has to be longer to cover them up (hi, joking)

in any case, you can solve statement 1 with numerical intuition, if you have that sort of intuition.

Yes, but in this case I think the brute-force method is still straight forward and safe. Human brain (at least my brain) easily mistakes illusion for intuition.

I'm just curious: Which is an effective way to develop intuition for something, such as PS or CR or RC?

My friend asks me:
- Have you ever dreamt about solving a math question?
- No. I just dreamt of something like eating (so I have an intuitive sense about which food is good for me)
- Do you have good math background?
- No.
- … head shaking… (hopeless)
- But I have a super intuitive Teacher, who can pull solutions down from the air…. Haha, now you tell me!

[NL]

the pink stuff makes me think you're saying "i can stop here because i'm thinking 'not sufficient'."

99% correct. Correct one: I can stop here because it’s time for a conclusion. Then I got a wrong conclusion.

if that's the case, then, no you can't "stop at this point", since both of these results give the same value for z (= the actual goal of the problem).

If these results give the same value for z, so you can stop because it’s sufficient (may be you made mistake with typo)

[RonPurewal]

If these results give the same value for z, so you can stop because it’s sufficient (may be you made mistake with typo)

^^ no.
if you get the same two values for z, you still need to make sure that no other values are possible.

this is, in fact, the entire reason why your earlier question--"how do i know that those are the only values of u^2 and v^2?"--is relevant (and crucial).