# Thread: Math proof in AP or GP

1. ## Math proof in AP or GP

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.

2. 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

3. 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:

$4(10^{k}+2) \sum_{n=1}^k(10^{n-1})+1$

We can substitute this:

$\sum_{n=1}^k10^{n-1} = \frac{10^{k}-1}{9}$

and then we see...

$4(10^{k}+2) \sum_{n=1}^k(10^{n-1})+1 = \frac{ (5^{k}2^{k+1})(5^{k}2^{k+1} + 2)+1}{9} = \left [\frac{5^{k}2^{k+1}+1}{3}\right] ^{2}$

Done.. yes?

Now, how did I get this: $4(10^{k}+2) \sum_{n=1}^k(10^{n-1})+1$?

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.

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

Call the terms in the sequence $(a_0)^2, \; (a_1)^2, \; (a_2)^2, \; \ldots$ with $(a_0)^2 = 49, \; (a_1)^2 = 4489, \; (a_2)^2 = 444889$, and so on.

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

$a_n \; = \; 1 \; + \; \sum_{k=0}^{n}6 \cdot 10^k$ (1)

We also notice that:

$(a_n)^2 \; - \; (a_{n-1})^2 \; = \; 44 \cdot 10^{2n} \; + \; 4 \cdot 10^n$ (2)

(we have to increase the last 4 by 4, and prepend "44" to each $(a_n)^2$ to get to the next one)

Now we apply conjecture (1):

$(a_n)^2 \; - \; (a_{n-1})^2 \; = \; \left[ a_{n-1} \; + \; 6 \cdot 10^n \right]^2 \; - \; (a_{n-1})^2$

$= \; 36 \cdot 10^{2n} \; + \; 12 \cdot 10^n \cdot a_{n-1}$

$= \; 36 \cdot 10^{2n} \; + \; 12 \cdot 10^n \; + \; \sum_{k=n}^{2n-1} 72 \cdot 10^k$

$= \; 36 \cdot 10^{2n} \; + \; 14 \cdot 10^n \; + \; \sum_{k=n+1}^{2n-1} 2 \cdot 10^k \; + \; \sum_{k=n+1}^{2n-1} 7 \cdot 10^k \; + \; 7 \cdot 10^{2n}$

$= \; 43 \cdot 10^{2n} \; + \; 4 \cdot 10^n \; + \; 10^{(n+1)} \; + \; \sum_{k=n+1}^{2n-1} 9 \cdot 10^k$

$= \; 43 \cdot 10^{2n} \; + \; 4 \cdot 10^n \; + \; 10^{2n}$

$= \; 44 \cdot 10^{2n} \; + \; 4 \cdot 10^n$

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...

5. Originally Posted by Absane
and then we see...

$4(10^{k}+2) \sum_{n=1}^k(10^{n-1})+1 = \frac{ (5^{k}2^{k+1})(5^{k}2^{k+1} + 2)+1}{9} = \left [\frac{5^{k}2^{k+1}+1}{3}\right] ^{2}$

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

$5^{k}2^{k+1}+1 = 2 \cdot 10^k + 1$

This makes it more visible that the sum of the digits is 3, so we know $5^{k}2^{k+1}+1$ is divisible by 3 and $\frac{5^{k}2^{k+1}+1}{3}$ is an integer.

6. 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?

7. Originally Posted by przyk
I don't know if I'm stating the obvious, but you should probably add that:

$5^{k}2^{k+1}+1 = 2 \cdot 10^k + 1$

This makes it more visible that the sum of the digits is 3, so we know $5^{k}2^{k+1}+1$ is divisible by 3 and $\frac{5^{k}2^{k+1}+1}{3}$ 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:

$4(100^{k}+10^{k}) = (3d)^{2} - 1 = (3d-1)(3d+1)$

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

8. Originally Posted by 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?
The squares are the numbers in the sequence {49, 4489, 444889, ... }. I took them as squares of numbers (the an'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 $444889 - 4489 = 440400$. If we can show that $667^2 - 67^2$ is also $440400$, and we happen to know that $67^2 = 4489$, then it follows that $667^2 = 444889$.

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.

9. Originally Posted by przyk
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.

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