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Topic: A Cubic Arithmetic Sequence Puzzle (Read 635 times) |
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K Sengupta
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A Cubic Arithmetic Sequence Puzzle
« on: Dec 27th, 2006, 5:24am » |
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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.
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| « Last Edit: Dec 27th, 2006, 5:24am by K Sengupta » |
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K Sengupta
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Re: A Cubic Arithmetic Sequence Puzzle
« Reply #1 on: Dec 31st, 2006, 12:05am » |
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I append hereunder the proposed solution to the foregoing problem as follows:
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.
However, I have been unable to deduce a shorter ( but comprehensive methodology) giving all possible solutions whenever q is less than 1000.
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| « Last Edit: Jan 2nd, 2007, 10:02am by K Sengupta » |
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ThudnBlunder
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Re: A Cubic Arithmetic Sequence Puzzle
« Reply #2 on: Dec 31st, 2006, 8:06pm » |
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q2(q - 6a) = 2(a3 - 1)
When q = 6a + 2 we get the Mordell Curve
q2 = a3 - 1
which has no non-trivial solutions.
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