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Any others? (Posted on 2013-06-18) Difficulty: 2 of 5
100 is a square which is also a sum of 4 consecutive cubes.
Are there any others?

No Solution Yet Submitted by Ady TZIDON    
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No Subject | Comment 7 of 9 |
Start with Jer's factoring: y^2 = 2(2x+3)(x+3x+6).  

If d = greatest common divisor of (2x+3) and (x+3x+6), it also divides 4 times the second expression minus the square of the first expression = 15.  So d=1,3,5,15.  The following handles d=1.

Since (2x+3) can't be twice anything, set 

(2x+3) = a^2 and (x^2+3x+6) = 2b^2

Solve the first for x and substitute in the second to get

(a^4+15) = 8b^2 = 2c^2 for c = 2b 

So (2c^2 - 15) is the square of a integer.

If c = 0 or 5 mod 10 then (2c^2 - 15) = 85 or 35 mod 100 which is impossible.

Other values of c mod 10 give (2c^2 - 15) = 3 or 7 mod 10 which is also impossible for squares.

  Posted by xdog on 2013-06-21 10:59:46
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