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        riddles >> medium >> A Cubic Arithmetic Sequence Puzzle
        
(Message started by: K Sengupta on Dec 27^th, 2006, 5:24am)

Title: A Cubic Arithmetic Sequence Puzzle
Post by K Sengupta on Dec 27^th, 2006, 5:24am

Consider three positive integers p< q< r in arithmetic sequence.

Determine analytically all possible solutions of the equation:
p^3 + q^3 = r^3 - 2, whenever q is less than 116.

Title: Re: A Cubic Arithmetic Sequence Puzzle
Post by K Sengupta on Dec 31^st, 2006, 12:05am

I append hereunder the proposed solution to the foregoing problem as follows:

[hide]

SOLUTION :

Let p = q-a and r = q+a

Then,p^3 + q^3 = r^3 - 2 yields:
q^2(q-6a) = 2(a^3-1)...(i)
Since, LHS must be divisible by 2, it follows that q is even. So substituting q = 2s, we obtain:
4*s^2(s-3a)+ 1 = a^3
Or, a^3 = 1 (Mod 4)
Hence, q is even and a is odd....(ii)

Now, if q<6a then, LHS of (i) is negative, while RHS of (i) is non-negative. This is a contradiction.
Hence, q>=6a.

Case A: q = 6a

From (i), we obtain:
a^3 = 1, giving a=1, so that q=6, yielding:
(p, q, r) = (5,6,7)

Case B: q is greater than 6a

Minimum value of q is (6a+1), so that:
(6a+1)^2 <= q^2(q-6a)= 2(a^3-1)
So, 2*a^3 >= 36*a^2 + 12a +3 > 36*a^2
Or, a^3> 18*a^2, giving:
a>18, so that a>=19, yielding:
q> 6a+1>= 115, so that q>=116 as q must be even

Consequently, (p, q, r) = (5,6,7) is the only possible solution whenever q is less than 116.[/hide]

However, I have been unable to deduce a shorter ( but comprehensive methodology) giving all possible solutions whenever q is less than 1000.

Title: Re: A Cubic Arithmetic Sequence Puzzle
Post by THUDandBLUNDER on Dec 31^st, 2006, 8:06pm

q²(q - 6a) = 2(a³ - 1)
When q = 6a + 2 we get the Mordell Curve
q² = a³ - 1
which has no non-trivial solutions.