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Flying Numbers
2005-10-11 15:06:37

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.

MathsIsFun
2005-10-10 16:58:29

That's really good, ganesh!

ganesh
2005-10-10 14:31:22

A simple proof I can think of is
(a+b+c)²=a² + b² + c² + 2ab + 2bc + 2ac
a=10000, b=200, c=1
The resultant is 100000000 + 40000 + 1 + 4000000 + 400 + 20000 = 104060401.
This will continue as n increases.
For example, for n=2,
a=1000000, b=2000, c=1.
The resultant would be 1000000000000 + 4000000 + 1 + 4000000000 + 4000 + 2000000 = 1004006004001.
It can be seen that the other digits are not affected because of the number of zeros in a and b.
Hence, for any value of n, the resultant is a perfect square.
q.e.d

Flying Numbers
2005-10-10 08:26:45

Hello all,
I've been working on this proof and can't seem to get anywhere with it.... any help would be appreciated!

Between every two digits of the number 14641, n zeros are inserted. For example:

n=1 ; 104060401
n=2 ; 1004006004001
etc.....

Prove that the obtained numbers for n greater than or equal to 1 are all perfect squares.

So far I can see that the square root of the obtained numbers will be:
n=1; 10201
n=2; 1002001
n=3; 100020001
etc....

Therefore by a simple case by case analysis I can see that the obtained values are all perfect squares because their square roots are integers....

I'm just really unsure as to how I should go about proving this. I was thinking on the lines of a proof by contradiction (similar to the way that you prove sqrt(2) is irrational), but I'm stuck. Thanks in advance!!