Simple proof about squares!

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   

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

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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? 

Hi jsaddern and catmg,     

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)