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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.

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## Comments

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TopNewestUsual 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.

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Thanks sir.

But how do you find x - y

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

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

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National Mathematics Talent Contests

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@Nihar Mahajan @Swapnil Das @Calvin Lin @Niranjan Khanderia @Satyajit Mohanty

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

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