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Another Quickie (Posted on 2011-03-13) Difficulty: 2 of 5
Let n and a be positive integers such that 3n^2=4a-1.

Prove that a is a difference of consecutive cubes.

See The Solution Submitted by broll    
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Solution spoiler | Comment 1 of 2

4a-1 is ODD and divisible by 3


 We can  assume 4a-1= 3*(2*k+1)^2=12*k^2+12*k+3


4*a=4*(3*k^2+3k+1)

a=3*k^2+3k+1= (k+1)^3-k^3  qed

Edited on March 25, 2017, 9:10 pm
  Posted by Ady TZIDON on 2011-03-13 16:31:14

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