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| Cubic Diophantine Conclusion (Posted on 2016-04-22) |
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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
A ffine problem (spoiler)
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 | Comment 7 of 9 | 
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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).
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Posted by Harry
on 2016-05-02 07:06:03 |
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