# 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
4 years, 3 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

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.

- 4 years, 3 months ago

Thanks sir.

But how do you find x - y

- 4 years, 3 months ago

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

- 4 years, 3 months ago

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

- 4 years, 3 months ago

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

- 4 years, 3 months ago

Sir, are you professor?

- 4 years, 3 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 :/

- 4 years, 3 months ago

- 4 years, 3 months ago

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

- 4 years, 3 months ago