Solve for positive integers x, y & z :

\(x + y - z = 4\)

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

\(xyz = 6 \)

Solve for positive integers x, y & z :

\(x + y - z = 4\)

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

\(xyz = 6 \)

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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. – Satyajit Mohanty · 1 year, 11 months ago

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But how do you find x - y – Dev Sharma · 1 year, 11 months ago

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– Satyajit Mohanty · 1 year, 11 months ago

\((x-y)^2 = (x+y)^2 - 4xy\)Log in to reply

@Dev Sharma : Is this any of NMTC's problems? Nowadays, NMTC is trending among the young students of India! – Satyajit Mohanty · 1 year, 11 months ago

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– Dev Sharma · 1 year, 11 months ago

sir, it RMO problem. What is full form of NMTC?Log in to reply

National Mathematics Talent Contests – Satyajit Mohanty · 1 year, 11 months ago

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– Dev Sharma · 1 year, 11 months ago

Sir, are you professor?Log in to reply

– Satyajit Mohanty · 1 year, 11 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 :/Log in to reply

I made it to a bi quadratic in z but no integral sol. – Aakash Khandelwal · 1 year, 11 months ago

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@Nihar Mahajan @Swapnil Das @Calvin Lin @Niranjan Khanderia @Satyajit Mohanty – Dev Sharma · 1 year, 11 months ago

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