Math proof in AP or GP

[Nivvedan]

Prove that the numbers 49, 4489, 444889, 44448889,..... obtained by inserting 48 into the middle of the preceding mumbers are squares of integers.
Please give a formal mathematical proof involving Arithmetic progression or Geometric progression, not just verification.

[andbna]

First, we shall take the square root of the example numbers

sqrt(49)=7,sqrt(4489)=67, sqrt(444889)=667, sqrt(44448889)=6667... This can be expressed as: 1+(the sum of (6*[10^i]) from i=0 to i=n)
So for 7, n=0, for 67 n=1, etc...
which we can then express as: 1 + 6(10^(n+1)-1)/9
that squared (ie (1+6(10^(n+1)-1)/9)^2 ) should equal the following series:

49, 4489, 444889, 44448889... can be expressed as: 1+(the sum of (8*(10^i)) for all i from 0 to n)+(the sum of (4*(10^[i+n+1])) for all i from 0 to n)

which can be expressed as: 1 + 8*(10^(n+1)-1)/9 + 40*(10^n*(10^(n+1)-1))/9

Thus, for all integers of n when n>=0

1 + 8*(10^(n+1)-1)/9 + 40*(10^n*(10^(n+1)-1))/9 = (1+6(10^(n+1)-1)/9)^2

easiest way to make sure we are correct at this stage is to use a GDC, which does confirm that the equations are equivilent (or at least, for what values it graphed, meaning it is still technicaly a conjecture right now.)

now, we will simply have to reduce each side to being the same...

1 + 8*(10^(n+1)-1)/9 + 40*(10^n*(10^(n+1)-1))/9 = (1+6(10^(n+1)-1)/9)(1+6(10^(n+1)-1)/9)
1 + 8*(10^(n+1)-1)/9 + 40*(10^n*(10^(n+1)-1))/9 = 1 + 2(6(10^(n+1)-1)/9) + 6(10^(n+1)-1)/9)^2 //expand a bit
8*(10^(n+1)-1)/9 + 40*(10^n*(10^(n+1)-1))/9 = 12(10^(n)*10-1)/9) + (6(10^(n+1)-1)/9))^2 //subtract 1 from each side
(8*10^(n+1)/9) - 8/9 + 40*10^n*(10^(n+1)-1)/9 = (12*10^(n+1)/9) - 12/9 + ((6*10^(n+1)/9)-6/9)^2 //Play with numbers...
(80*10^n/9) - 8/9 + 400*10^(2n)/9 - 40*10^n/9 = 120*10^(n)/9 - 12/9 + ((6*10^(n+1)/9)-6/9)^2
(80*10^n/9) - 8/9 + 400*10^(2n)/9 - 40*10^n/9 = 120*10^(n)/9 - 12/9 + 4/9 - 8*10^(n+1)/9 + (6*10^(n+1))^2/81 //expanding
80*10^n - 8 + 400*10^(2n) - 40*10^n = 120*10^(n) - 12 + 4 - 8*10^(n+1) + (6*10^(n+1))^2/9 //multiply by 9
40*10^n - 8 + 400*10^(2n)= -8 + 40*10^(n) + (6*10^(n+1))^2/9 //Collect like terms
400*10^(2n)=(6*10^(n+1))^2/9 //add 8, subtract 40*10^n
3600*10^(2n)=(6*10^(n+1))^2 //multiply by 9
3600*10^(2n)=6*10^(n+1)*6*10^(n+1)
3600*10^(2n)=36*10^(n+1)*10^(n+1)
3600*10^(2n)=36*10*10^(n)*10*10^(n)
3600*10^(2n)=3600*10^(n)*10^(n)
10^(2n)=10^(n)*10^(n)

and clearly,

10^(2n)=10^(2n)

Thus the conjecture is proven.

Sorry about the lack of that fancy TEXT stuff or w/e it's called
-Andrew

[Absane]

Ok Andrew. I didn't read what you put.. but by the looks of it, that doesn't count. That's not to say that what I am going to post will fly either... but it's cleaner and can easily be made into a formal proof.

Ok, I figured out that the kth term of the sequence {49, 4489, 444889, ...} is given by:



We can substitute this:



and then we see...



Done.. yes?



Now, how did I get this: ?

Well, I noted this pattern:

49 = 4*10 + 9
4489 = 44*100 + 8*10 + 9
444889 = 444*1000 + 88*10 + 9
...

Then you see a pattern to derive a function. From there, you just collect like terms and simplify.

[przyk]

A less elegant (I'm no mathematician) method, since we're all giving away the answer:

Call the terms in the sequence with , and so on.

Now we whip out a calculator, find the square roots of the first few terms, and conjecture:

(1)

We also notice that:

(2)

(we have to increase the last 4 by 4, and prepend "44" to each to get to the next one)

Now we apply conjecture (1):

This is the same result as (2), so it follows by induction that (1) indeed gives the roots of our original sequence. As luck would have it, they're all integers...

[przyk]

and then we see...



Done.. yes?

I don't know if I'm stating the obvious, but you should probably add that:

This makes it more visible that the sum of the digits is 3, so we know is divisible by 3 and is an integer.

[Absane]

przyk, maybe it's too early in the morning for me but I am having trouble seeing the logic of your proof. Are you doing induction? Why the difference of squares?

[Absane]

I don't know if I'm stating the obvious, but you should probably add that:

This makes it more visible that the sum of the digits is 3, so we know is divisible by 3 and is an integer.

You're right, but when I was doing the problem factoring 10 made it easier for me to find the factors of the following:



Factoring out 2 and 5 and permuting them in certain ways eventually led me to solve for d, the square factor for the problem. Of course, I could have then just reduced it more so that it's not that much of an eyesore. For that, I apologize

[przyk]

przyk, maybe it's too early in the morning for me but I am having trouble seeing the logic of your proof. Are you doing induction? Why the difference of squares?

The squares are the numbers in the sequence {49, 4489, 444889, ... }. I took them as squares of numbers (the a_n's) that we need to show are integers - specifically, {7, 67, 667, ... }.

The logic goes something like this (using the second and third terms to illustrate):

We know that . If we can show that is also , and we happen to know that , then it follows that .

In general, if I can show that the difference between two successive terms in the sequence {49, 4489, 444889, ... } is the same as the difference between the squares of the two corresponding terms in the sequence {7, 67, 667, ... }, and I know 7² = 49, I know I've correctly guessed all the roots.

[Absane]

I know I've correctly guessed all the roots.

Heh... well that's how a lot of proofs work. Out of nowhere, many start with something like "Let x be something very complicated that we won't derive for you..."

Heh.