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Prove that:

\((10^{n+1}+1)(a_n \cdot 10^{n} + a_{n-1} \cdot 10^{n-1} + \ldots + a_0 \cdot 10^0)\) is never a perfect square. (Where \(a_0, a_1, \ldots a_n\) are positive integers between \(0\) and \(9\).)

 

I can't seem to develop an argument using prime factorizations... I wonder what approach you'll use?

 

Thank you so much!

 Mar 12, 2022
 #1
avatar+117104 
+2

This is not a proof but I have convinced myself that the statement is true for n =1 to 9

 

\((10^{n+1}+1)>(a_n \cdot 10^{n} + a_{n-1} \cdot 10^{n-1} + \ldots + a_0 \cdot 10^0)\)

 

So if the product of the 2 is a perfect square then   \((10^{n+1}+1)\)    must have at least one factor that is a perfect square

 

I looked at \((10^{n+1}+1)\)    for n = 1 to 9       [And I have just realized that that restriction is not valid - there is no restriction placed on n]

 

anyway,  none of those have a perfect square factor so the statement is true (at least for n = 1 to 9)

 

NOT VERY HELPFUL - YES I KNOW THIS   angry

 Mar 12, 2022
 #2
avatar+117104 
+2

I think I can disprove it with an example.   n=10    (Edit correction, I had previously said n=11)

 

10^11+1 = 121*826446281

 

so

\(\sqrt{(10^{11}+1)*(826446281)} = \sqrt{121^2*826446281^2}=121*826446281 \)

 

\((10^{n+1}+1)(a_n \cdot 10^{n} + a_{n-1} \cdot 10^{n-1} + \ldots + a_0 \cdot 10^0)\\ when\;\; n=10\\ (10^{11}+1)(826446281)\qquad \text{is a perfect square}\)

 

 

sqrt((10^11+1)*826446281 = 9090909091

 

 

-------------------------------

 

Again I reread your question, it says 'between 0 and 9'  Is that supposed to include 0 and 9?

My answer uses 0.

I suppose I wasn't meant to but is 9 also really not allowed? 

 Mar 12, 2022
edited by Melody  Mar 13, 2022
 #3
avatar+2401 
+1

I was also quite confused at the case when n = 10. 

I believe your example fits all the requirements. 

I'm pretty sure between 0 - 9 includes 0 and 9. 

 

=^._.^=

catmg  Mar 12, 2022
 #4
avatar+72 
+1

Yes, it does include \(0\) and \(9\).

 

Also, yes! It looks like \(11\) does disprove it. Can you describe the set of all numbers for which the given expression is a perfect square? (Maybe to get there, how did you come up with your example?)

jsaddern  Mar 13, 2022
 #5
avatar+117104 
+2

Hi jsaddern and catmg,     laugh

 

I will try and explain how I was thinking.

 

I know this.

\((10^{n+1}+1)>(a_n \cdot 10^{n} + a_{n-1} \cdot 10^{n-1} + \ldots + a_0 \cdot 10^0)\)

 

So I know that if    \((10^{n+1}+1)(a_n \cdot 10^{n} + a_{n-1} \cdot 10^{n-1} + \ldots + a_0 \cdot 10^0)\)   is a perfect sqare then

 

\((10^{n+1}+1)\)     must have at least one perfect square factor.   Can you see that?

 

So let         \((10^{n+1}+1) =a^2b\)         where a and b are rational positve numbers

then   \((a_n \cdot 10^{n} + a_{n-1} \cdot 10^{n-1} + \ldots + a_0 \cdot 10^0)\)    must be     bc^2   where b and c are rational positive numbers

and  a>c

That would make    \((10^{n+1}+1)(a_n \cdot 10^{n} + a_{n-1} \cdot 10^{n-1} + \ldots + a_0 \cdot 10^0)= a^2b*bc^2=(abc)^2\)

 

So I googled online for a number of the form  10^n+1  with a perfect square factor.

I found this site:

https://math.stackexchange.com/questions/1188750/prove-that-all-numbers-10n-1-are-square-free

which said that    10^11 +1  has a factor of 11^2

 

then I simply divided   (10^11+2) by  121 to get the other factor.

 

(10^11+1)/121 = 826446281

so    (10^11+1)*826446281 = 11^2 * 826446281^2 which is a perfect square

 

826446281*4 = 3 305 785 124                   so      (10^11+1)* 3 305 785 124    is anothother one

3305785124*9 = 29 752 066 116               so      (10^11+1)*  29 752 066 116  is another one    

3305785124*16 = 52 892 561 984             so      (10^11+1)*  52 892 561 984  is another one.

52892561984*25 = 1 322 314 049 600        TOO big 

 

So now I have found four examples that show the statement is not true.

 

I hope that all makes sense to you.  (hopefully there is no flaw in my logic)    laugh

 Mar 13, 2022
edited by Melody  Mar 13, 2022
edited by Melody  Mar 13, 2022
 #6
avatar+72 
+2

Thank you! I profoundly appreciate your response.

 

I understand your logic and see the four counter-examples you provide. I will continue to work on describing all numbers that work as counter examples. Thanks Melody,

 

jsaddern

jsaddern  Mar 14, 2022
 #7
avatar+117104 
+2

All counter examples will be of the form  

a^2b=bc^2    

as I have described above.

There is most likely an infinite number of them.

Melody  Mar 14, 2022
 #8
avatar+2401 
+1

Your logic makes perfect sense to me. :))

I think that if 10^(n+1) + 1 contains at least 1 perfect square factor, then this statement can be disproven. 

 

=^._.^=

catmg  Mar 14, 2022

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