Diophantine sequence
Dean Hickerson
dean at math.ucdavis.edu
Thu Oct 17 23:35:28 CEST 2002
Ed Pegg wrote:
> 2^n - 2 = 7(x^2 - x) + (y^2 - y)
>
> Euler proved that the above diophantine equation has a unique solution
Actually a unique *positive* solution; you can change x to 1-x and/or y to
1-y.
> for any n.
For any positive n.
> (Engel, Problem-Solving Strategies) He never published his result or proof.
>
> This leads to an odd set of sequences:
> {1, 2, 3, 1, 6, 5, 7, 16, 3, 29, 34, 24, 91, 44, 138, 225, 51, 500, 399,
> 601, 1398}
> {1, 1, 1, 2, 1, 3, 4, 2, 9, 6, 12, 23, 1, 46, 45, 47, 136, 43, 229, 314,
> 144}
Note that the equation is equivalent to
2^(n+2) = (2y-1)^2 + 7 (2x-1)^2,
so it's related to norms of elements of the ring of integers in the quadratic
field Q(sqrt(-7)), and Euler's claim presumably follows from unique
factorization in that field. (I haven't checked the details.)
From this we can get a formula for the x's and y's: Let a(n) and b(n)
be the unique rational numbers such that
a(n) + b(n) sqrt(-7) = ((1 + sqrt(-7))/2)^n.
I.e.
a(n) = (((1 + sqrt(-7))/2)^n + ((1 - sqrt(-7))/2)^n)/2
and
b(n) = (((1 + sqrt(-7))/2)^n - (((1 - sqrt(-7))/2)^n))/(2 sqrt(-7)).
Then x(n) = abs(b(n)) + 1/2 and y(n) = abs(a(n)) + 1/2:
n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
y(n) 1 2 3 1 6 5 7 16 3 29 34 24 91 44 138 225 51 500 399
x(n) 1 1 1 2 1 3 4 2 9 6 12 23 1 46 45 47 136 43 229
Dean Hickerson
dean at math.ucdavis.edu
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