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!

jsaddern Mar 12, 2022

#1**+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

Melody Mar 12, 2022

#2**+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?

Melody Mar 12, 2022

#5**+2 **

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)

Melody Mar 13, 2022

#6**+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