Sum of all pairs

find sum of all possible x+y where x and y are natural number solutions of the equation

 
 x^3 + y^3 = x^2 + y^2 + 42xy

?
In general this is a really hard problem (Diophantine equation)... I'm sure that if this is a real question from somewhere, there must be some trick to solving it, but I'm not sure what it is off the top of my head.

https://en.wikipedia.org/wiki/Diophantine_equation

I would start just by plotting the equation (iterate over x = -50 to 50, y = - 50 to 50 or something like that) and see if there's any sort of pattern. Of course, just graphical inspection doesn't prove that there isn't a really far-out number that also solves the equation.

It seems pretty chaotic to me, I don't see a clear pattern.
https://www.wolframalpha.com/input/?i=plot++round%28x%5E3+%2B+y%5E3+-+x%5E2+-+y%5E2+-+42+x+y%29+%3D+0+from+x+%3D+-40+to+40%2C+y+%3D+-40+to+40

EDIT: I didn't realize you said natural numbers. I was thinking all integers!

The equation looks clearly bounded by graphical inspection on the +x, +y quadrant.
https://www.wolframalpha.com/input/?i=plot++round%28x%5E3+%2B+y%5E3+-+x%5E2+-+y%5E2+-+42+x+y%29+%3D+0+from+x+%3D+0+to+50%2C+y+%3D+0+to+50

What I said still applies -- just graphical inspection doesn't prove it, but it looks pretty solid.

So all you need to do is iterate over a bounded range of positive {x, y} pairs, see which ones solve the equation.

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Since we're just working with positives, it becomes clear after I stare at the equation for a bit that the x^3 + y^3 term will eventually beat the x^2 + y^2 + 42xy term, but initially the RHS is greater than the LHS due to the large constant factor (42). Still that's no where near a proof.
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60

1<=x,y<=31 (strictly, sqrt(x^2+y^2)<=22.sqrt(2), and that's provable for the first quadrant, as here), so it's not many to search.
(1,7),(7,1),(22,22)
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But where did you calculate the 22 from?

Edit: To clarify, I understand that (22,22) happens to be the furthest point out radially in the 1st quadrant, but I can't think of an easy way to actually calculate that without knowing beforehand that that is the furthest point out.
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Ganado wrote:
But where did you calculate the 22 from?


Use polar coordinates.

Substitute
x = r cos θ
y = r sin θ
into the equation.

Divide by r2 (which is not zero in the present circumstances) and you will get (after a minor trig identity) the polar equation of the curve
r = ( 1 + 21 sin(2θ) ) / ( cos3 θ + sin3 θ )

In the first quadrant the numerator has a maximum and the denominator has a minimum fortuitously at the same angle (π/4 radians, or 45 deg). At this point you get the maximum radius in the first quadrant:
r = 22*sqrt(2) = 31.something
(This actually corresponds to one solution, x=y=22).

So search for all integer points inside or on a sector of radius 31 (or, more conveniently, just all x and y less than or equal to 31.)

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I see, so that shows that the radius is bounded, and therefore you only have to search within the maximum of that polar function. That is such a nicer way than what I was trying to do. Thanks.
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