1,2

Euler (unpublished) showed there is a unique positive solution (x,y) for every positive n.

Proof of uniqueness: Let R be the ring of integers of Q(sqrt(-7)) and let a=(1+sqrt(-7))/2, b=(1-sqrt(-7)/2. It is easy to see that any element of R of even norm (=squared absolute value) can be divided by one of a or b to get back an element of R. Thus since ab=2, the only elements in R of norm 2^n and of the form (p+q*sqrt(-7))/2 with p,q odd are a^n, b^n, -a^n, -b^n - precisely one of which lies in the first quadrant. Finally apply Dean Hickerson's remarks. - Paul Boddington (psb(AT)maths.warwick.ac.uk), Jan 23 2004

A. Engel, Problem-Solving Strategies, Springer-Verlag, New York, 1998.

Note that the equation is equivalent to 2^(n+2) = (2y-1)^2 + 7 (2x-1)^2, so it is 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. 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. - Dean Hickerson (dean.hickerson(AT)yahoo.com), Oct 19, 2002.

a(n)=2^(n/2)*abs(cos(n*t))+1/2, where t=arctan(sqrt(7)). - Paul Boddington (psb(AT)maths.warwick.ac.uk), Jan 23 2004

(PARI) p(n, x, y)=2^n-2-7*(x^2-x)-(y^2-y) a(n)=if(n<0, 0, y=1; while(frac(real(component(polroots(p(n, x, y)), 2)))>0, y++); y)

Cf. A076632.

Sequence in context: A114576 A116468 A110237 this_sequence A035485 A074306 A036039

Adjacent sequences: A076628 A076629 A076630 this_sequence A076632 A076633 A076634

nonn,easy

Ed Pegg Jr. (ed(AT)mathpuzzle.com), Oct 17 2002

More terms from Benoit Cloitre (benoit7848c(AT)orange.fr), Oct 24 2002

More terms from Lambert Klasen (lambert.klasen(AT)gmx.de), Jan 14 2005