OFFSET
1,2
COMMENTS
Euler (unpublished) showed there is a unique positive solution (x,y) for every positive n.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 a*b=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, Jan 23 2004
REFERENCES
A. Engel, Problem-Solving Strategies, Springer-Verlag, New York, 1998.
FORMULA
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, Oct 19 2002
a(n) = 2^(n/2)*abs(cos(n*t))+1/2, where t=arctan(sqrt(7)). - Paul Boddington, Jan 23 2004
MATHEMATICA
Table[Reduce[{2^n-2==7(x^2-x)+(y^2-y), x>0, y>0}, {x, y}, Integers][[-1, -1]], {n, 50}] (* Harvey P. Dale, Dec 15 2018 *)
PROG
(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)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Ed Pegg Jr, Oct 17 2002
EXTENSIONS
More terms from Benoit Cloitre, Oct 24 2002
More terms from Lambert Klasen (lambert.klasen(AT)gmx.de), Jan 14 2005
Definition corrected by Harvey P. Dale, Dec 15 2018
STATUS
approved