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 Quarting the cube (Posted on 2013-02-28)

Select integer x and triangular number y such that 8y=3x^4-2x^2-1.

Prove that y is divisible by 28 - or find a counter-example.

 See The Solution Submitted by broll Rating: 5.0000 (2 votes)

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 Solution - no cube spotted | Comment 2 of 6 |
Since y is triangular let y = t(t+1)/2
8t(t+1)/2 = 3x^4-2x^2-1
t^2 + t - (3x^4-2x^2-1)/4 = 0
t = (-1 ± √(1 + 3x^4-2x^2-1))/2
t = (-1 ± x√(3x^2 -2))/2

What is then sought is to find integers for x that make t an integer, so we need 3x^2 - 2 to be a perfect square
http://oeis.org/A001835 has the formula

a(n) = 4*a(n-1) - a(n-2), with a(0)=1, a(1)=1
Terms are the solutions to: 3x^2-2 is a square

If you take these numbers and put them into √(3x^2 -2) you get x
http://oeis.org/A001834 has the formula

b(0) = 1, b(1) = 5, b(n) = 4b(n-1) - b(n-2)
(I changed a to b to tell them apart.)

So what is sought is to prove t is an integer now that
t = (1 ± a(n)b(n))/2
By inspection a(n) is always odd and b(n) is always odd so this is clearly true.

 Posted by Jer on 2013-03-01 09:22:59

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