Solving A Nice Polynomial System by Math26039335

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This problem comes from a fellow mathematician on twitter. Thank you Math26039335!!!
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[ssalmero]

Using substitution: x+y=a, x-y=b in the second equation gets ab^2=45, so the only solution is a=5 and b=+-3, then {4, 1} and {1, 4} are the solutions.

[Camarelli]

If we write x²+y²=[(x+y)²+(x-y)²]/2 and (x²-y²)=(x-y)(x+y) we can immediatly write the system in terms of u=x+y and v=x-y, quickly finding that u=5 and v=±3 and then the (x, y) solutions (4, 1) and (1, 4).

[jmart]

Substracting the two equations you find:
2xy(x+y)=40
The first eq can be transformed adding and substracting 2xy.
(x+y)(x^2+y^2+2xy-2xy) = 85
(x+y)((x+y)^2-2xy) = 85
(x+y)^3-2xy(x+y) = 85
Then substituting the first result
(x+y)^3 = 125
x+y = 5
And substituting this result in the first eq
xy = 4

From which you obtain
x=4, y=1 or viceversa.

[dennisdesormier]

It's not OCD — it's style. :-)
Speaking of style, you substituted at 7:49, true to form. It does factor directly as (4x – y)(x – 4y) = 0, skipping a couple lines to get to y = 4x and x = 4y. That factorization looks kind of cool (at least to me).

[randomjin]

Note that if (x, y) is a solution then (y, x) is too. Set x = kcoshz, y = ksinhz, then by the hyperbolic identities: k³eᙆ(e²ᙆ+e⁻²ᙆ) = 170, k³ = 45eᙆ, so 45e²ᙆ(e²ᙆ+e⁻²ᙆ) = 170 meaning e⁴ᙆ = 25/9, so eᙆ = √5/√3 and thus k = √15; getting back to exponential forms of sinh and cosh we get x = (√15(√5/√3+√3/√5))/2 = 4, y = (√15(√5/√3-√3/√5))/2 = 1. Hence (x, y) = (4, 1) is a solution as is (1, 4)

[Alphabet]

the blue curve at the end (x+y)(x^2+y^2)=85 is not quite an elliptic curve, as one can find by taking the coordinate transform a=x+y b=x-y, but it is projectively equivalent to one as it is a nonsingular cubic curve. the same should also be true of the orange curve for similar reasons

[shmuelzehavi]

Another approach:
The given system of equations is:
(x + y) (x^2 + y^2) = 85 (1)
(x - y) (x^2 - y^2) = 45 (2)
( and we can easily see that the equations are symmetric in x and y )
Let u = x + y, v = x - y and the system becomes:
u (u^2 + v^2) = 170 (3)
uv^2 = 45 (4)
By subtracting eq. (4) from eq. (3) we obtain:
u^3 = 125 = 5^3
Therefore, if we restrict ourselves to real numbers, then u = 5, v = ± 3 and we get:
x = 4, y = 1 or x = 1, y = 4.
If we extend to complex numbers we get:
u^3 = 125 = 5^3 exp⁡(2inπ), where n = 0, ±1 and therefore:
u = x + y = 5 exp⁡(2inπ/3) (5)
v = x - y = 3m exp(-inπ/3), where m = ±1 (6)
Therefore:
x = ((5 exp(2inπ/3) + 3m exp(-inπ/3)) / 2
y = ((5 exp(2inπ/3) - 3m exp(-inπ)/3) / 2
where, m = ±1 and n = 0, ±1, giving us 6 distinct complex solutions.
For n=0 we get the 2 real pairs: (x, y) = (4, 1) and (x, y) = (1, 4)

[rajeshbuya]

Awsum sum with a beautiful graph. Thanks to your friend as well!

[chaosredefined]

Let s = x+y and d = x-y. Note that s^2 = x^2 + 2xy + y^2 and d^2 = x^2 - 2xy + y^2. So, (s^2 + d^2)/2 = x^2 + y^2. Also note that sd = x^2 - y^2. With this, our original equations are now:
s(s^2 + d^2)/2 = 85 => s^3 + s(d^2) = 170
d(sd) = 45 => s (d^2) = 45

Subtract these equations, and we get s^3 = 125. Therefore, s = 5. Substituting this into the second equation gives us 5 (d^2) = 45, which means that d^2 = 9, or d = 3 or -3.

If s = x+y = 5 and d = x-y = 3, then s + d = (x+y) + (x-y) = 2x, and s + d = 8. So, 2x = 8, or x = 4. Then, since x+y = 5, this means y = 1. So, (4, 1) is a potential solution, and checking with the original equations, this holds.

If s = x+y = 5 and d = x-y = -3, then s + d = (x+y) + (x-y) = 2x, and s + d = 2. So, 2x = 2, or x = 1. Then, since x+y = 5, this means y = 4. So, (1, 4) is a potential solution, and checking with the original equations, this holds.

So, two solutions exist (1, 4) and (4, 1)

[broytingaravsol]

multiplying these 2 equations side by side leads to 3 factors with 2 perfect squares, corresponding to 9 and 25 as the factors in RHS

[unique]

Thanks .Good to learn about treatment of homogeneous equations

[Kvakvazyabr]

Always loved your solutions 'cause their kind of elegant. Let me thank you for the interesting videos and offer my solution.

We can solve it by using substitution x = a + b and y = a - b.
Then the original equations become:
(a + b + a - b)*((a + b)^2 + (a - b)^2) = 85
(a + b - a + b)*((a + b)^2 - (a - b)^2) = 45
Simplify to:
4*a^3 + 4*a*b^2 = 85
8*a*b^2 = 45
Then, substituting the second equation into the first we get
4*a^3 + 45/2 = 85 -> a^3 = 125/8 -> a = 2.5
and
8*2.5*b^2 = 45 -> b^2 = 45/20 -> b = +/- 1.5
And last:
x = 2.5 +/- 1.5 and y = 2.5 -/+ 1.5 -> solutions are (1, 4) and (4, 1)

I, personally, love this substitution, because it helped me a lot in many similar equations

[mcwulf]

Nice problem. I found both those methods too, although my division was the other way round.

When you get the quadratic in X and y surely it's simple enough to factorise that without the k substitution.

[mystychief]

first equation: (x+y)*(x*x+y*y)=5*17. So integer solutions, x+y=5, x*x+y*y=17. Then you already see the solution 1, 4 and 4, 1. It fits the second equation. The only problem left is to prove, that these are the only solutions, which is probably easy because of the 5 and the 17 must fit and in the other one you have the 45 = 3*3*5.

[scottleung]

I got both solutions - BTW the graphs are awesome!

[rakenzarnsworld]

17*5=85
15*3=45
Case 1:
x+y=5, x^2+y^2=17, x-y=3, x^2-y^2=15
-> x=4, y=1

[pedrovargas]

I rearranged the second into (x+y)(x-y)² = 45, then compared this to the first.
85 = (5)(17) = (-5)(-17)
45 = (5)(9) = (-5)(-9).
No divisions there.
(x, y) = (4;1); (1;4); (-4, -1); (-1, -4).

[teambellavsteamalice]

Before watching the video (in seconds):
If there are integer solutions you get 5*17 and 3*15 as the only obvious solutions. Giving y=1 and thus x=4.

Let's see if I'm right!
Edit: almost, I didn't get the reverse, -3*-15 is also 45.
I should have seen it, as the parts for both equations can be swapped...

[MrLidless]

These are cubics, so you’re missing a solution. x = -³√i, y = -4 ³√i

[yoav]

When i solve syber's problems i feel yay i solved it!, but when syber solves his problems he says ok it's nice, but let's solve this with another method 😃💯