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## #1 2005-10-10 08:26:45

Flying Numbers
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### Perfect Square Proof

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!!

Last edited by Flying Numbers (2005-10-10 08:43:44)

## #2 2005-10-10 14:31:22

ganesh
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### Re: Perfect Square Proof

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

Character is who you are when no one is looking.

## #3 2005-10-10 16:58:29

MathsIsFun

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### Re: Perfect Square Proof

That's really good, ganesh!

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

## #4 2005-10-11 15:06:37

Flying Numbers
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### Re: Perfect Square Proof

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.