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 A Sum of Consecutive Square Roots (Posted on 2016-08-10)
Show every positive integer power of 1+sqrt(2) can be expressed as a sum of square roots of consecutive integers.

(1+sqrt[2])^n = sqrt(x) + sqrt(x+1)

 No Solution Yet Submitted by Brian Smith No Rating

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It is mainly a matter of putting the recurrences in a form where they can be manipulated.

Small values of x:

1
8
49
288
1681
9800
57121
332928
1940449
11309768

If a(n) is ODD: the lesser number (x) is the square of {1,7,41,239,1393,...}, and the greater (x+1) is twice the square of {1,5,29,169,985,...}. These give sqrt x = 1/2 ((1+sqrt(2))^(2n+1)-(1+sqrt(2))^(-(2n+1))) and sqrt (x+1) = 1/2((1+sqrt(2))^(2n+1)+(1+sqrt(2))^(-(2n+1))); adding these gives:

1/2 ((1+sqrt(2))^(2n+1)-(1+sqrt(2))^(-(2n+1)))+1/2((1+sqrt(2))^(2n+1)+(1+sqrt(2))^(-(2n+1))) = (1+sqrt(2))^(2n+1)

If a(n) is EVEN: the greater number (x+1) is the square of {3,17,99,577,3363,...}, and the lesser is twice the square of {2,12,70,408,2378,...}. These give sqrt (x+1) = ((1+sqrt(2))^(2*n)+(1-sqrt(2))^(2*n))/2 and sqrt (x) =  1/2 ((1+sqrt(2))^(2n)-(1-sqrt(2))^(2n)); adding these gives:

((1+sqrt(2))^(2*n)+(1-sqrt(2))^(2*n))/2+1/2 ((1+sqrt(2))^(2 n)-(1-sqrt(2))^(2 n)) = (1+sqrt(2))^(2n)

Other formulations of the defining equations are possible; it doesn't matter much as long as a consistent scheme is adopted between the pairs.

 Posted by broll on 2016-08-10 08:48:31

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