The Hidden Square - 24/04/2013

[showmyiq]

Look closely at this number:

x = 11....122....25

We know that the count of the 1s in this number is exactly n.
We know that the count of the 2s in this number is exactly n+1.
So we can conclude there are total n+n+1+1=2n+2 digits in this number.

Can you prove x is a square number?


*By square number we define such number that can be represented by a square
If A is a square number, some other integer y exists such as A=y*y
first square numbers: 1,4,9,16,25,36,49,etc.

[showmyiq]

The problem was solved by Sergey Sokolov via the facebook.

His method is attacking the given number strings with constructing An number with total N digits equal to 1. I have translated his solution using LaTeX and you can see it here:

\begin{equation}

A_n = \underbrace{111...1}_{n} \\

9*A_n + 1 = 9 * \underbrace{111...1}_{n} + 1 = \underbrace{999...9}_{n} + 1 = \underbrace{1\underbrace{000...0}_{n}}_{n+1} \\

(9*A_n + 1) * A_n = 1\underbrace{000...0}_{n} * \underbrace{111...1}_{n} = \underbrace{111...1}_{n}\underbrace{000...0}_{n} \\

(9*A_n + 1) * A_n + A_n = \underbrace{111...1}_{n}\underbrace{000...0}_{n} + \underbrace{111...1}_{n} =
\underbrace{111...1}_{2n} \\

(9*A_n + 1) * A_n + A_n + 10*A_n = \underbrace{111...1}_{2n} + \underbrace{111...1}_{n}0 = \underbrace{111...1}_{n-1}\underbrace{222...2}_{n}1 \\

(9*A_n + 1) * A_n + A_n + 10*A_n + 4 = \underbrace{111...1}_{n-1}\underbrace{222...2}_{n}1 + 4 =
\underbrace{111...1}_{n-1}\underbrace{222...2}_{n}5 \\

9*A_n^2 + 12 *A_n + 4 = \underbrace{111...1}_{n-1}\underbrace{222...2}_{n}5 \\

(3*A_n + 2)^2 = \underbrace{111...1}_{n-1}\underbrace{222...2}_{n}5 \\

\end{equation}

[showmyiq]

I want to add mine solution to this problem as well, using another approach:

\begin{equation}

\underbrace{111...1}_{n}\underbrace{222...2}_{n+1}5 = \\
10^2(\underbrace{111...1}_{2n} + \underbrace{111...1}_{n}) + 25 = \\
\frac {10^2} {9} ( 10^{2n} + 10^n -2) + 25 = \\
\frac {1} {9} ( 10^{2(n+1)} + 10^{n+2} - 200) + 25 = \\
\frac {1} {9} ( 10^{2(n+1)} + 10^{n+2} + 25) = \\
[\frac {1} {3}(10^{n+1} + 5)]^2 = \eta^2 \\
\text{for some natural number } \color{red}\eta

\end{equation}

[kudrik]

\begin{equation} \underbrace{111...1}_{n}\underbrace{222...2}_{2n}5 = \\
10^{3n}+...+10^{2n+1}+2*10^{2n}+...+2*10+5 = \\
10^{2n}*(10^n+...+10)+2*10^n*(10^n+...+10)+2*(10^n+ ... +10)+5 = \\
(10^{2n}+2*10^n+2)*(10^n+ ... +10)+5 = \\
(10^{2n} + 2*10^n + 2)*(10*(10^n - 1)/(10-1)) + 5 = \\
(10^{3n+1} + 2*10^{2n+1} + 2*10^{n+1} - 10^{2n+1} - 2*10^{n+1} - 20 + 45)/9 \\
(10^{3n+1} + 2*5*10^{2n} + 25)/9 \\
\text{only square if }3n+1=4n => n=1
\end{equation}

[showmyiq]

I just want to add that kudrik solution is based on my wrong publication on FB – I have made a typo that the count of the 2s is 2n. He proved that if the count of the 2s is exactly 2n, then we can receive a square number only if n is equal to 1.

Thank you kudrik for this solution!

I have a question about the 4th row.

First about the 2nd row the last bracets should be 10^n + ... + 10 ( I believe you have missed the dots notation).

But my real question is : how we receive from 10^n + 10 = 10*(10^n-1)/(10 -1) on the 4th row?
Maybe I am missing something?

[kudrik]

I've missed the dots twice
On the 4th row we have a sum of geometric progression (10^n + ... + 10)

[showmyiq]

Oh, thank you for the clarification!

For all the others who are wondering why we received this from 3th row to the 4th row I want to add this extra information:

If you have a geometric progression and a is the first term, n is the number of the terms and s is the step we use in order to get to the next term, the formula is:

\begin{equation}
SUM = a *\frac {1-s^n} {1-s}
\end{equation}

In our case we have 10 + ... + 10^n, so the first term is a=10, the number of the terms is exactly n, and the iteration step is exactly 10. So we have

\begin{equation}
SUM = a *\frac {1-s^n} {1-s} \\
SUM = 10 * \frac {1-10^n} {1-10}\\
SUM = 10 * \frac {10^n -1} {10-1} \\
SUM = \frac {10^{n+1} -10} {9}
\end{equation}

Ok, now I continue to the next 5th row

[showmyiq]

OK, I do agree about the last row.

\begin{equation}
(10^{3n+1} + 2*5*10^{2n} + 25)/9 \\
\end{equation}

My first thought here is to prove that this is natural number for every n.
Indeed the sum of the digits of this number, for every n>0 will be 27.
27 is dividable by 9, equal 3, so the whole number will be a square number, only if the nominator is a square number (because the denominator is a square number 9 equals 3^2).

Ok, so far – so good!

But why we do the assumption that the only suitable value is only when 4n=3n+1?
I do agree that 4n is excellent square number in the given circumstances:

\begin{equation}
10^{4n} + 2*5*10^{2n} + 25 = (10^{2n} + 5)^2 \\
\end{equation}

But why we conclude that 4n is the only suitable value?
Can we prove the other values will not hit some random square?

I have just written this simple C# code to analyze the issue further.
Imagine we have this similar equation to our:

\begin{equation}
2^{i+2} +2*2^i*1 + 1 \\
\end{equation}

If we use the same approach we should conclude that the only square number here is when i+2 = 2i, or i=2 is the only solution. But check this out:

i=0,Sum=7,SQRT=2.64575131106459
i=1,Sum=13,SQRT=3.60555127546399
i=2,Sum=25,SQRT=5
i=3,Sum=49,SQRT=7
i=4,Sum=97,SQRT=9.8488578017961
i=5,Sum=193,SQRT=13.8924439894498
i=6,Sum=385,SQRT=19.6214168703486
i=7,Sum=769,SQRT=27.7308492477241
i=8,Sum=1537,SQRT=39.2045915678253
i=9,Sum=3073,SQRT=55.4346462061408
i=10,Sum=6145,SQRT=78.390050389064
i=11,Sum=12289,SQRT=110.855762141623
i=12,Sum=24577,SQRT=156.770532945449
i=13,Sum=49153,SQRT=221.704758631835
i=14,Sum=98305,SQRT=313.536281792076
i=15,Sum=196609,SQRT=443.406134373443
i=16,Sum=393217,SQRT=627.070171511929
i=17,Sum=786433,SQRT=886.810577293708
i=18,Sum=1572865,SQRT=1254.13914698489
i=19,Sum=3145729,SQRT=1773.62030885982
i=20,Sum=6291457,SQRT=2508.27769594995
i=21,Sum=12582913,SQRT=3547.24019485571
i=22,Sum=25165825,SQRT=5016.55509288994
i=23,Sum=50331649,SQRT=7094.48017827945
i=24,Sum=100663297,SQRT=10033.1100362749
i=25,Sum=201326593,SQRT=14188.9602508429
i=26,Sum=402653185,SQRT=20066.2199977973
i=27,Sum=805306369,SQRT=28377.9204488278
i=28,Sum=1610612737,SQRT=40132.4399582183
i=29,Sum=3221225473,SQRT=56755.8408712266
i=30,Sum=6442450945,SQRT=80264.8798977486
i=31,Sum=12884901889,SQRT=113511.681729239
i=32,Sum=25769803777,SQRT=160529.759786153
i=33,Sum=51539607553,SQRT=227023.36345187
i=34,Sum=103079215105,SQRT=321059.519567634
i=35,Sum=206158430209,SQRT=454046.726900437
i=36,Sum=412316860417,SQRT=642119.039132932
i=37,Sum=824633720833,SQRT=908093.453799222
i=38,Sum=1649267441665,SQRT=1284238.0782647
i=39,Sum=3298534883329,SQRT=1816186.90759762

It appears that i=3 is another solution too!

Here is the source to see what is going under the results:

using System;

class Program
{
static void Main()
{
for (long i = 0L; i < 40; i++)
{
long S = Convert.ToInt64(Math.Pow(2, i + 2)) + 2 * 1 * Convert.ToInt64(Math.Pow(2, i)) + 1;
double check = Math.Sqrt(S);
Console.WriteLine("i={0},Sum={1},SQRT={2}",i,S,check);

}
}
}

I think the probability to hit a square number in future is close to 0, but can we say it is equal to 0?

[kudrik]

\begin{equation} 2^{i+2} +2*2^i*1 + 1 \\ \end{equation}
is not equal to
\begin{equation} (10^{3n+1} + 2*5*10^{2n} + 25)/9 \\ \end{equation}

n=3 will not give you a square number: sqrt(1112222225) = 33349.995877

[showmyiq]

Yes, they are not equal at all – I just gave it as example.

I am still researching this topic, because I think it’s even more interesting than the original math problem! I will explain it further with details.

When we reached the last equation:
\begin{equation} (10^{3n+1} + 2*5*10^{2n} + 25)/9 \\ \end{equation}

It’s obvious that the nominator should be a square number in order the whole expression to be a square number. So we have this equation we need to analyze is it a square number and if yes – for which values of n.
\begin{equation} 10^{3n+1} + 2*5*10^{2n} + 25 \\ \end{equation}

The question disturbing me since then is – “Are we sure that n=1 is the only possible solution?”

I gave you an example in which an equation with the same pattern appears to have more than 1 possible solution.
\begin{equation} 2^{i+2} +2*2^i*1 + 1 \\ \end{equation}

In this case we have the same power equation with 2 variable elements and 1 constant element (the same structure as the original equation). And in this case we have minimum 2 possible solutions – i=2 and i=3. How can be sure some n>1 doesn’t exists in which this equation:
\begin{equation} 10^{3n+1} + 2*5*10^{2n} + 25=0 \\ \end{equation}

is fulfilled for some integer n?

Let’s define a natural number M such that:

\begin{equation}
10^{3n+1} + 2*5*10^{2n} + 25 = M^2
\\ \end{equation}

As you can see all the elements in the left side of this equation are dividable by 5. Therefore if such M exists it should be dividable by 5. So let’s say M = 5*M1.

\begin{equation}
10^{3n+1} + 2*5*10^{2n} + 25 = M^2 \\
10^{3n+1} + 2*5*10^{2n} + 25 = (5M_1)^2 \\
10^{3n+1} + 2*5*10^{2n} + 25 = 25M_1^2 \\
2^{3n+1}5^{3n+1} + 10^{2n+1} + 25 = 25M_1^2 \\
2^{3n+1}5^{3n+1} + 2^{2n+1}5^{2n+1} + 25 = 25M_1^2 \\
2^{3n+1}5^{3n-1} + 2^{2n+1}5^{2n-1} + 1 = M_1^2 \\
\\ \end{equation}
Therefore we can conclude that M1 is an odd number, so let’s say M1=2*M2 + 1, for some random natural number M2.

\begin{equation}
2^{3n+1}5^{3n-1} + 2^{2n+1}5^{2n-1} + 1 = M_1^2 \\
2^{3n+1}5^{3n-1} + 2^{2n+1}5^{2n-1} + 1 = (2*M_2 + 1)^2 \\
2^{3n+1}5^{3n-1} + 2^{2n+1}5^{2n-1} + 1 = 4*M_2^2 + 4*M_2 + 1 \\
2^{3n+1}5^{3n-1} + 2^{2n+1}5^{2n-1} = 4*M_2^2 + 4*M_2 \\
2^{3n-1}5^{3n-1} + 2^{2n-1}5^{2n-1} = M_2^2 + M_2 \\
2^{2n-1}5^{2n-1} ( 2^n5^n + 1 ) = M_2(M_2 +1) \\
\\ \end{equation}
Now let’s analyze the last equation, so I can explain my thoughts further.
If we say that n=1 is the only possible solution, we in fact say that M2 is equal to 10^(2n-1) is the only possible case. How can be sure about that?

Let’s continue with our thoughts. Let’s say M2 is dividable by 2. Can M2+1 dividable by 2 too? No!
Even if M2 is dividable by some random integer random number A, you can easily deduct that M2 +1 will be not dividable by the same random A (no matter what is the value of A, M2+1 divided by A will give us reminder of 1). So, Either M2 is dividable by 2, or M2 +1 is dividable by 2. Let’s say M2 is dividable by 2, then M2 should be equal to M3*2^{2n-1}, for some random natural number M3.

\begin{equation}
2^{2n-1}5^{2n-1} ( 2^n5^n + 1 ) = M_2(M_2 +1) \\
M_2 = M_3*2^{2n-1} \\
2^{2n-1}5^{2n-1} ( 2^n5^n + 1 ) = M_3*2^{2n-1}(M_3*2^{2n-1} +1) \\
5^{2n-1} ( 2^n5^n + 1 ) = M_3*(M_3*2^{2n-1} +1) \\
\\ \end{equation}
And we should not forget that M3 should not be dividable by 5. We can continue with many more substitutions and we will not reach contradictions (just the possible n solution will rise and rise). So my question is can you prove that n=1 is the possible solution? Why in the case I gave we have 2 solutions, not only 1 (i=2 and i=3). The most interesting question is - how to predict the behavior of the equation when we are dealing with power equations? Because I have reached to M19 following this logic and I am still in circle. I have reached this equation at the end:

\begin{equation}
2^m + 1 = 5^n
\\ \end{equation}
Does this equation has other solution than m=2 and n=1 (if m and n are natural numbers)?
At general – how to solve such equations with powers and 2 variables?