# Crocodile System

Solve for positive integers x, y & z :

$$x + y - z = 4$$

$$x^2 - y^2 + z^2 = 4$$

$$xyz = 6$$

Note by Dev Sharma
2 years, 8 months ago

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Usual Method:

There're no integral solutions. As $$(x,y,z)$$ will only be the 9 permutations of $$(1,1,6)$$ and $$(1,2,3)$$. None of them satisfy.

Brute Force Method:

I also did some nonsense calculations to find the values:

$$x+y = 4+z, xy=\dfrac{6}{z}$$

Thus I obtained $$x-y = \sqrt{\dfrac{z^3+8z^2+16z-24}{z}}$$

$$(x+y)(x-y) = 4 - z^2 \Rightarrow (x+y)^2(x-y)^2 = (4 - z^2)^2$$

$$\Rightarrow (4+z)^2 \left(\dfrac{z^3+8z^2+16z-24}{z} \right)= (4 - z^2)^2$$

And then visit this.

No integral solutions obtained by Brute Force as well.

- 2 years, 8 months ago

Thanks sir.

But how do you find x - y

- 2 years, 8 months ago

$$(x-y)^2 = (x+y)^2 - 4xy$$

- 2 years, 8 months ago

@Dev Sharma : Is this any of NMTC's problems? Nowadays, NMTC is trending among the young students of India!

- 2 years, 8 months ago

sir, it RMO problem. What is full form of NMTC?

- 2 years, 8 months ago

Sir, are you professor?

- 2 years, 8 months ago

As you can see, I'm only 18 years old. So, I'm a student, not a professor :D And please don't call me Sir. Sounds weird :/

- 2 years, 8 months ago

I made it to a bi quadratic in z but no integral sol.

- 2 years, 8 months ago