+72 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!
+118687 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 
+118687 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?
+118687 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)
+72 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
