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Square+27 (Posted on 2019-01-22) Difficulty: 3 of 5
Find all positive integers x > y > z such that x2+27y, y2+27z, z2+27x and x2+y2+z2+27 are all perfect squares.

No Solution Yet Submitted by Danish Ahmed Khan    
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Not the way to it | Comment 1 of 2
I don't know how to do this. It's surely not with the approach I've taken here. 

I have done 

y=(2x+27b)*b
z=(2y+27c)*c

Then I have arbitrarily give b=c=1 and calculate x so that z^2+27x is a perfect square. 

For x=165 y=357 z=741, I get

x^2+27y=192^2

y^2+27z=384^2

z^2+27x=744^2

unfortunately x^2+y^2+z^2+27 is not a perfect square. I can give different values to a and b and search again for new posible values of x, but would be time consuming... 

So, this is probably not the way to do this. 



  Posted by armando on 2019-01-23 06:48:24
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