Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**Flying Numbers****Member**- Registered: 2005-10-09
- Posts: 2

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 (2005-10-09 10:43:44)*

Offline

**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 23,087

A simple proof I can think of is

(a+b+c)²=a² + b² + c² + 2ab + 2bc + 2ac

When n=1 in your case,

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

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,660

That's really good, ganesh!

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

Offline

**Flying Numbers****Member**- Registered: 2005-10-09
- Posts: 2

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.

Offline

Pages: **1**