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Quartic Quandary (Posted on 2011-08-23) Difficulty: 3 of 5
Consider three positive integers x < y < z in arithmetic sequence, and determine all possible triplets (x,y,z) that satisfy this equation:

x4 + x2y2 = z4 76

No Solution Yet Submitted by K Sengupta    
Rating: 3.0000 (1 votes)

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Solution Comment 1 of 1

I If (x-a)^2*x^2+76 = (x+a)^4 -(x-a)^4, then

II x(8a^3-a^2 x+10ax^2-x^3)=76, with x={1,2,4,19,38,76},

when {a=1,x=2} is the only solution in the positive integers.


  Posted by broll on 2011-08-24 06:39:28
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