Math Is Fun Forum

Hello,
Thanks for the replies, I just worked it out using a different method.
I wrote the number (14641) as:

k=0: 1*10^4 + 4*10^3 + 6*10^2 + 4*10 + 1 ----------- (=14641)
k=1: 1*10^8 + 4*10^6 + 6*10^4 + 4*10^2 + 1
k=2: 1*10^12 + 4*10^9 + 6*10^6 + 4*10^3 + 1
k=n: 1*10^(4n+4) + 4*10^(3n+3) + 6*10^(2n+2) + 4*10^(n+1) + 1

and I wrote the square roots of the above numbers as:

k=0: 1*10^2 + 2*10 + 1 ---------- (=121)
k=1: 1*10^4 + 2*10^2 + 1
k=2: 1*10^6 + 2*10^3 + 1
k=n: 1*10^(2n+2) + 2*10^(n+1) + 1

It's easy to see that the square root for k=n is an integer, since n can only be an integer, so now it's just necessary to prove that these forumals derived by the patterns are in fact correct. To do this I used induction to prove that:

SQRT (1*10^(4n+4) + 4*10^(3n+3) + 6*10^(2n+2) + 4*10^(n+1) + 1) = 1*10^(2n+2) + 2*10^(n+1) + 1

I first checked the base, which works.... then I assumed n=k to be true and checked n=k+1. I obtained equivalent expressions and so I showed that by induction the formula i derived for the squareroot when n=k is infact true, and this is also an integer.... making the original expression a perfect square.

This is kind of long, but does anyone follow. I really like ganesh's way of proving this, very nice.