Question

Find an expression for the continual radical C=√m+√m+√m+...
in terms of of m that does not involve a continued radical. Then determine all positive integers m so that C is a positive integer.

Answer 1

square both sides to get rid of the first square root:

C^2=m+(√m+√m+√m+...)

we already know that √m+√m+√m+... =C, so substitute that in:

C^2=m+C

You should be able to solve it from that :)

Answer 2

you need to clarify if this is:\[1) C = \sqrt{m}+\sqrt{m}+...\]or is it:\[2) C=\sqrt{m+\sqrt{m+\sqrt{m+...}}}\]

Answer 3

number 2

Answer 4

in that case @Zara has given you the correct start on this problem. You basically get left with:\[C^2-C-m=0\]which is a quadratic equation in C.

Answer 5

if you solve this quadratic using the formula for quadratic equations, you get:\[C=\frac{1\pm\sqrt{1+4m}}{2}\]since C has to be a positive integer, we can eliminate the solution which "1 - ...", giving:\[C=\frac{1+\sqrt{1+4m}}{2}\]this implies "1+4m" must be a perfect square, so we can write:\[1+4m=n^2\qquad\text{(for some n)}\]this can be re-arranged to:\[m=\frac{n^2-1}{4}=\frac{(n-1)(n+1)}{4}\]since m must also be an integer, that mean the numerator must be a multiple of 4. if n is even, then the numerator is always an odd number times another odd number, which will result in some other odd number. therefore n MUST be ODD and can therefore be written as:\[n=2k+1,\qquad\text{for k=1,2,3,...}\]substituting this into the equation for m gives:\[m=\frac{(2k+1-1)(2k+1+1)}{4}=\frac{2k(2k+2)}{4}=\frac{4k(k+1)}{4}=k(k+1)\]we therefore have a solution for m given by:\[m=k(k+1),\qquad\text{for k=1,2,3...}\]
you can test a few values out, e.g.:
k=1 gives m=2 and C=2
k=2 gives m=6 and C=3

Answer 6

I just noticed that k can start from zero to given the trivial solution:
k=0 gives m=0 and C=0

Answer 7

hope this helps...

Answer 8

Thank you sooooo much... I realy appreciate it

Answer 9

yw - it was an interesting problem to solve :-)