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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
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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

Mar 12, 2022
#2
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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?

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

Mar 12, 2022
edited by Melody  Mar 13, 2022
#3
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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
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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?)

#5
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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)

Mar 13, 2022
edited by Melody  Mar 13, 2022
edited by Melody  Mar 13, 2022
#6
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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,

#7
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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
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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