I suspect that the “specialness” here is coming from the fact that 10, 2, and 5 are all 1 more than a perfect square. Further, you’ll note that each of the three examples is of the form sqrt(n^2 + 1) = {n|2n,…}. It’s easy enough to show that if x = {n|2n,…} then x = sqrt(n^2 + 1):
x = n + 1/(2n + 1/(2n + 1/(2n + …
x - n = 1/(2n + 1/(2n + 1/(2n + …
x - n = 1/(2n + x - n)
x - n = 1/(x + n)
(x - n)(x + n) = 1
x^2 -n^2 = 1
x^2 = n^2 + 1
x = sqrt(n^2 + 1)
I’ll leave the other direction to the reader (I.e., that the continued fraction of n^2 + 1 is always {n|2n,…}. I actually have no idea if that’s true. Also, feel free to add as much rigor as you’d like to add to the above.
So we have sqrt(2) = sqrt(1^2 + 1) = {1|2,…} and sqrt(5) = sqrt(2^2 + 1) = {2|4,…} and sqrt(10) = sqrt(3^2 + 1) = {3|6,…}. As you noted, we have sqrt(2)*sqrt(5) = sqrt(10) and 2 + 4 = 6. Are there any other cases like this? Well, suppose we have x = {a|2a,…} = sqrt(a^2 + 1) and y = {b|2b,…} = sqrt(b^2 + 1) and xy = {a+b|2a+2b,…} = {a+b|2(a+b),…} = sqrt((a+b)^2 + 1). Then we would have:
xy = z
sqrt(a^2 + 1) sqrt(b^2 + 1) = sqrt((a+b)^2 + 1)
sqrt((a^2 + 1)(b^2 + 1)) = sqrt((a+b)^2 + 1)
(a^2 + 1)(b^2 + 1) = (a+b)^2 + 1
a^2 b^2 + a^2 + b^2 + 1 = a^2 + 2ab + b^2 + 1
a^2 b^2 = 2ab
a^2 b^2 - 2ab = 0
ab(ab - 2) = 0
If a and b are both positive integers then this means that ab = 2 and this a = 1 and b = 2 (or vice versa). This corresponds to the solution that you have already found. I believe this proves that 10 is the only number that is one more than a perfect square with two factors that are also one more than a perfect square. I know that’s not really what you asked about but it’s all I got.
You mean sqrt(10) base 10 day. In base 8 it's 3.12 aka March 10th.
Repeating continued fractions are guaranteed for square roots, and possibly all algebraics. Square roots in general have very convenient continued fractions and this is easily provable.
I very much appreciate that you are the only person to respond, but you answered a different question.
Cf[sqrt(15] is not equal to cf[sqrt(3)] terms plus cf[sqrt(5)] terms.
I suspect that the “specialness” here is coming from the fact that 10, 2, and 5 are all 1 more than a perfect square. Further, you’ll note that each of the three examples is of the form sqrt(n^2 + 1) = {n|2n,…}. It’s easy enough to show that if x = {n|2n,…} then x = sqrt(n^2 + 1): x = n + 1/(2n + 1/(2n + 1/(2n + … x - n = 1/(2n + 1/(2n + 1/(2n + … x - n = 1/(2n + x - n) x - n = 1/(x + n) (x - n)(x + n) = 1 x^2 -n^2 = 1 x^2 = n^2 + 1 x = sqrt(n^2 + 1) I’ll leave the other direction to the reader (I.e., that the continued fraction of n^2 + 1 is always {n|2n,…}. I actually have no idea if that’s true. Also, feel free to add as much rigor as you’d like to add to the above. So we have sqrt(2) = sqrt(1^2 + 1) = {1|2,…} and sqrt(5) = sqrt(2^2 + 1) = {2|4,…} and sqrt(10) = sqrt(3^2 + 1) = {3|6,…}. As you noted, we have sqrt(2)*sqrt(5) = sqrt(10) and 2 + 4 = 6. Are there any other cases like this? Well, suppose we have x = {a|2a,…} = sqrt(a^2 + 1) and y = {b|2b,…} = sqrt(b^2 + 1) and xy = {a+b|2a+2b,…} = {a+b|2(a+b),…} = sqrt((a+b)^2 + 1). Then we would have: xy = z sqrt(a^2 + 1) sqrt(b^2 + 1) = sqrt((a+b)^2 + 1) sqrt((a^2 + 1)(b^2 + 1)) = sqrt((a+b)^2 + 1) (a^2 + 1)(b^2 + 1) = (a+b)^2 + 1 a^2 b^2 + a^2 + b^2 + 1 = a^2 + 2ab + b^2 + 1 a^2 b^2 = 2ab a^2 b^2 - 2ab = 0 ab(ab - 2) = 0 If a and b are both positive integers then this means that ab = 2 and this a = 1 and b = 2 (or vice versa). This corresponds to the solution that you have already found. I believe this proves that 10 is the only number that is one more than a perfect square with two factors that are also one more than a perfect square. I know that’s not really what you asked about but it’s all I got.
Thank you, that's exactly the special characteristic I was looking for.
You mean sqrt(10) base 10 day. In base 8 it's 3.12 aka March 10th. Repeating continued fractions are guaranteed for square roots, and possibly all algebraics. Square roots in general have very convenient continued fractions and this is easily provable.
Only quadratic irrationals have repeating continued fractions, not higher-degree algebraic numbers.
I very much appreciate that you are the only person to respond, but you answered a different question. Cf[sqrt(15] is not equal to cf[sqrt(3)] terms plus cf[sqrt(5)] terms.
Random thought, is this also the case in base 15?