Find the sum of all possible values of x² + y²

This problem is from the Harvard-MIT Mathematics Tournament.

[peternewseterforever]

What a video. It's really funny when you realize your mistakes, but at least you solve them. Also, good job at finding the solutions; they were quite surprising.

[ahpx]

Best math teacher i have ever seen, most think i love is your smile 😊

[Grecks]

Although not technically required to answer the problem, we can solve this equation system completely. Let's do it. Let S be the solution set (as a subset of R^2 with pairs (x, y) as elements).

We have 4 disjoint cases:
1) x = y = 0: Observe the following: If (x, y) is in S, then x = 0 iff y = 0. Also note that (0, 0) is in S. Therefore, in this case we have 1 solution (0, 0) and a corresponding x^2 + y^2 = 0.

2) 0 != x = y: The equation system reduces to x^3 = 4x and we have: ((x, x) in S) iff (x^2 = 4) iff (x = 2 or x = -2). Therefore, we have two distinct solutions in this case: (2, 2) and (-2, -2), and the corresponding value of x^2 + y^2 is 8.

3) 0 != x = -y: The system reduces to x^3 = 2x and we have: ((x, -x) in S) iff (x^2 = 2) iff (x = sqrt(2) or x = -sqrt(2)). Therefore, we also have two distinct solutions in this case: (sqrt(2), -sqrt(2)) and (-sqrt(2), sqrt(2)), and the corresponding value of x^2 + y^2 is 4.

4) x != 0 and y != 0 and |x| != |y|: This is the main case. It follows that x + y != 0 and x - y != 0. By adding and subtracting the two equations, we can conclude that x^3 + y^3 = 4(x + y) and x^3 - y^3 = 2(x - y) which is equivalent to x^2 - xy + y^2 = 4 and x^2 + xy + y^2 = 2 in this case (by dividing by (x + y) or (x - y) respectively). Adding and subtracting the last two equations to/from each other yields: 2(x^2 + y^2) = 6 and -2xy = 2 which is equivalent to x^2 + y^2 = 3 and xy = -1. So we already know the value of x^2 + y^2 in this case. But we can derive more: (x + y)^2 = 3 + 2(-1) = 1 and (x - y)^2 = 3 - 2(-1) = 5. This is equivalent to |x + y| = 1 and |x - y| = sqrt(5).

Now, whenever (x, y) is in S, we also have (y, x), (-x, -y), and (-y, -x) in S due to the symmetries of the original equation system. The important thing to note here is that in our case (4) all of these 4 solutions are pairwise distinct! Thus, all solutions that fall under this case come in groups of 4 distinct pairs. Therefore, without loss of generality, we can add the additional assumptions (0 < |x| < |y|) and (y > 0) to this case.

But are there any solutions in this case and, if yes, how many? Let's continue and see. First, observe that -- under the above additional assumptions -- (x + y) and (x - y) cannot have the same sign, because (x + y)(x - y) = |x|^2 - |y|^2 < 0. Second, note that x + y = x + |y| >= -|x| + |y| > 0 (under the above assumptions). Therefore we have: (x + y > 0) and (x - y < 0), and further: (|x + y| = x + y = 1) and (|x - y| = y - x = sqrt(5)).

By adding and subtracting the last two equations to/from each other, we arrive at: (2y = 1 + sqrt(5)) and (2x = 1 - sqrt(5)) which is equivalent to (x = -1/Phi) and (y = Phi) where Phi is the "golden ratio". It's easy to check that (x, y) = (-1/Phi, Phi) is indeed a solution.

Therefore, in this case we have 4 distinct solutions: (-1/Phi, Phi), (Phi, -1/Phi), (1/Phi, -Phi), and (-Phi, 1/Phi), and the corresponding value of x^2 + y^2 is 3.

--

In summary, the total number of distinct solutions (x, y) in S is: 1 + 2 + 2 + 4 = 9.

To answer the posed problem:
If we sum over all distinct values of (x^2 + y^2) for solutions (x, y) in S we get: 0 + 8 + 4 + 3 = 15 (this sum was most likely asked for in the problem).
If we sum the value of (x^2 + y^2) over all distinct solutions (x, y) in S we get: 1*0 + 2*8 + 2*4 + 4*3 = 36.

[criskity]

I love how you catch your mistakes and leave your lessons organic. Reminds me of my university professors!

[slavinojunepri]

13:20 - "and this, is what we've been looking for" with a look that gives goosebumps.😅

[cosmosapien]

We can use the two formulas:-
x³-y³=(x-y)(x²+y²+xy) and
x³+y³=(x+y)(x²+y²-xy)

when (x-y) and (x+y) are not 0. We can calculate separately for when they are 0.

Substituting values of x³ and y³ in the above formulas, we get

2 = x²+y²+xy and
4 = x²+y²-xy
which gives x²+y²=3

Now for the (x-y)=0 and (x+y)=0 cases, we can substitute x=y and x=-y respectively and get
x²+y² = 4 and
x²+y² = 8 respectively.

So total three possible values, and their sum is 3+4+8= 15

[Jeremy-id]

Thank you for your wonderful videos. Your love of maths and the natural way your body and personality express this are beautiful and your and our blessing.

For this problem, I suggest the last step of adding the two equations you did to get the final value of 3 for x^2 + y^2 needs some justification / expansion. This is because they were only cases - the alternatives to x=y and x=-y respectively - which do not have to be true simultaneously.

To avoid the need for this, rather than also factorise x^3 + y^3, I prefer to stick with the factorisation of x^3 - y^3 you have and solve the alternative case to x=y, ie the equation x^2 + y^2 + xy -2 = 0, directly. Dividing equation 1 by x and equation 2 by y and subbing for x^2 + y^2 gives 6 + y/x -+ x/y = 2 - xy. Adding the two fractions on LHS and again subbing for x^2 + y^2 leads to a quadratic equation in xy which factorises and has solutions -1 and -2.

The first of these values for xy gives the final value of 3 (= 2--1) for x^2 + y^2.

The second value for xy gives the value of 4 (2--2) for x^2 + y^2 you found from the x=~y case. This is because when y=~2/x is subbed into equations 1 and 2, it leads to two identical quadratic equations in x^2 and in y^2 which only have solutions for which x=-y.

So the first of the two equations you used in the last step can hold when x=~y as well as when the second equation you used holds. Nonetheless it is possible for both equations to be true simultaneously, leading to the final value of 3 for x^2 + y^2.

So with your approach, by justification / expansion, I guess I just mean this. Spelling out that because you have shown that A (x=y) or B (your first equation) is true, and also that C (x=-y) or D (your second equation) is true, 4 cases arise: A and C, A and D, B and C and B and D. And that by adding your two equations you are considering the last of these cases, the first three having already been dealt with, being those leading to the values of 0, 8 and 4 for x^2 + y^2 respectively.

Again, thank you for your lovely and inspirational videos and look forward to seeing the next one ❤

so much for your wonderful forward to your next video

[boguslawszostak]

for x<> 0 and y <> 0 we can substitute t=x/y
x^2=3+1/ t
y^2=3+t

So:
sum=x^2+y^2=6+t+1/t

x^2/y^2=t^2=(3+1/ t)/(3+t)
so
t^3*(3+t)=3t+1

t^4+3t^3-3t-1=0
(t^4-1)+3t(t^2-1)=0
(t^2+1)(t^2-1)+3t(t^2-1)=0


t=1 gives 6+1+1=8
t=-1 gives 6-1-1=4
(t^2+3t+1)=0 gives to solution witch sum is - 3 an product of roots is 1 so
t+1/t is sum of roots
gives us
sum= 6-3=3
0+2+4+8=15

[prakrit]

Respect to you Sir 🫡 The videos show thy love for the subject. Today is 'Guru Purnima' in India, a day to respect thy teachers 😇 So, once again, respect to you Sir for the informative videos, Thank you 😇😊

[franolich]

Very interesting problem. A variation on the solution:

x^3 = 3x + y [eq.1]
y^3 = x + 3y [eq.2]

Either x=y=0 or neither x nor y are 0 (x=0 => y=x^3-3x=0 etc). So in the case where both x and y are non-zero:

[eq.1] / x:
=> x^2 = 3 + y/x [eq.3]
[eq.2] / y:
=> y^2 = 3 + x/y [eq.4]

Let R = x^2 + y^2.

[eq.3] + [eq.4]:
=> R = 6 + y/x + x/y
=> R - 6 = (x^2 + y^2) / xy
=> (R-6)xy = R [eq.5]

[eq.2] - [eq.1]:
=> y^3-x^3 = 2(y - x)
=> (y-x)(x^2 + xy + y^2) - 2(y-x) = 0
=> (y-x)(x^2 + xy + y^2 - 2) = 0

Case 1: x = y
[eq.1] => x^3 = 4x
=> x = y = 0, ±2
=> R = 0, 8

Case 2: x^2 + xy + y^2 - 2 = 0
=> xy = 2 - R [eq.6]

[eq.6] in [eq.5]:
=> (R-6)(2-R) = R
=> 2R - R^2 - 12 + 6R = R
=> R^2 - 7R + 12 = 0
=> (R-4)(R-3) = 0
=> R = 3, 4

∴ R = x^2 + y^2 ∈ {0, 3, 4, 8}

[RyanLewis-Johnson-wqxs]

X^3=3X+Y Y^3=3Y+X X^2+Y^2=8 I don’t have to think about it.

[prime]

A Mathematican who is a good teacher. A rare combination!!

[domanicmarcus]

Sir, starting the time of 6:53, all the work is incorrect. You should have been solving x^2-xy+y^2-4 as equal to zero.

[SGuerra]

Muito bonita a questão! Parabéns pela escolha. Essa eu fiz facilmente. Brasil - agosto 2024. Very beautiful question! Congratulations on your choice. I did this easily. Brazil - August 2024.

[renyxadarox]

12:25 you can't conclude that x²+y²=3 is a solition without check x and y are both real.

[nanamacapagal]

BEFORE WATCHING:

I already suspect something with sum and difference of cubes. Let's add and subtract both equations and see what we get
x^3 + y^3 = 4x + 4y
x^3 - y^3 = 2x - 2y
Pull out some factors
(x+y)(x^2 - xy + y^2) = 4(x+y)
(x-y)(x^2 + xy + y^2) = 2(x-y)
Consider for the moment that x ≠ y and x ≠ -y.
x^2 - xy + y^2 = 4
x^2 + xy + y^2 = 2
Now add both equations
2x^2 + 2y^2 = 6
x^2 + y^2 = 3
Can this happen?
Subtract the equations instead.
2xy = -2
x^2 + 2xy + y^2 = 1
(x+y)^2 = 1
x^2 - 2xy + y^2 = 5
(x-y)^2 = 5
x and y can be real, this checks out

What about when x = y?
x^3 = 3x + x
x^3 = 4x
x^3 - 4x = 0
x(x-2)(x+2) = 0
x = 0, 2, -2
-> y = 0, 2, -2
x^2 + y^2 = 0, 8
Or when x = -y?
x^3 = 3x - x
x^3 = 2x
x^3 - 2x = 0
x(x-sqrt(2))(x+sqrt(2)) = 0
x = 0, sqrt(2), -sqrt(2)
y = 0, -sqrt(2), sqrt(2)
x^2 + y^2 = 0, 4
(0, 0) is a duplicate solution
Total: 3 + 0 + 8 + 4 = 15

[doug]

My approach was to get the equations into a form involving x^2 + y^2 from the beginning, so I realized that if I multiplied the first equation by x and the second equation by y and then subtracted the second from the first, I get x^4 - y^4 = 3x^2 + 3y^2. Then moving everything to one side and factoring, I get (x + y)(x - y)(x^2 + y^2 - 3) = 0. Then I solved x and y for when x = y and x = -y and also knew that x^2 + y^2 = 3 is also a solution. Is this a valid approach?

[joseluishablutzelaceijas]

Thanks for the video and the solution. Maybe it is worth quickly checking that or arguing why x^2 + y^2 = 3 also leads to real solutions for x and y? (this is anyways the case, and 3 is therefore a legitimate value for x^2 + y^2)

[Aenderson]

I make some calculus and another solutions are (φ, 1/φ), (1/φ, φ), (-φ, -1/φ) and (-1/φ, -φ), since φ=golden ratio((1+sqrt(5))/2)

[Burlongaming]

This is one of the easiest questions so far