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 Cubic Diophantine Conclusion (Posted on 2016-04-22)
Each of A, B and C is a positive integer that satisfies this equation:

A3 + B3 = 31C3

Find the smallest value of A+B+C

 No Solution Yet Submitted by K Sengupta Rating: 4.0000 (1 votes)

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 A ffine problem (spoiler) | Comment 7 of 9 |
A pity to leave this one without a derivation of armando’s findings.
Rather than invoke giants’ shoulders etc. here’s a school maths
approach which starts from armando’s first solution.

Since C is not zero, let x = A/C and y = B/C then A3 + B3 = 31C3

transforms to the cubic curve  x3 + y3 = 31  with a rational point

P(x1, y1) = P(137/42, -65/42) corresponding to armando’s result.

We need only find the point where the tangent at P meets the

curve again. It will also be rational and give us a second solution.

Gradient: dy/dx = -x2/y2, so the tangent at P(x1, y1) will have the

equation:    y = mx + c where     m = -(137/-65)2 = 18769/4225

and       c = -65/42 - m(137/42) = 54864/4225

Where it crosses the curve again,  x3 + (mx + c)3 = 31, which

simplifies to:     (m3 + 1)x3 +3m2cx2 + 3 mc2x + c3 – 31 = 0

We know there is a double root at P, (x = x1), so the LHS factors:

(m3 + 1)x3 +3m2cx2 + 3 mc2x + c3 – 31 = (m3 + 1)(x – x1)2(x – x2)

and by equating the constant terms, (or other coefficients), the

remaining root, x2, can be found:      c3 – 31 = (m3 + 1)(- x12x2)

giving   x2 = (31 – c3)/(x12(m3 + 1))

Substituting for m, c, and x1 :    x2 = 277028111/119531076.

Thus   y2 = 316425265/119531076 and these give another trio:

(A, B, C) = (277028111, 316425265, 119531076) as armando found.

In the same way, this new solution spawns other solutions, but will

they have A, B and C all positive and is there a smaller solution?

It’s worth drawing the cubic curve x3 + y3 = 31 and a few tangents. Its

shape provides some useful insights and a lot of fun.

(‘Rational Points on Elliptic Curves’ by Silverman & Tate was very useful
– many years ago).

 Posted by Harry on 2016-05-02 07:06:03

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