## Digit block repetition in prime squares

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• Considering squares with an odd number of digits, the cases where the squared integers are prime and the squares less than 10^9 appear to be only these (the
Message 1 of 7 , Aug 28, 2012
Considering squares with an odd number of digits, the cases where the squared integers are prime and the squares less than 10^9 appear to be only these (the repetition meaning the square looks like AbA, where A is a string of digits and b any digit):

121 = 11^2, 29929 = 173^2, 69169 = 263^2, 732947329 = 27073^2.

Does anyone know how uncommon these are for larger squares?

{Regarding the related question about squares with an even number of digits and a complete repetition AA - I couldn't find any solutions at all for "small" squares, whether or not the original numbers were prime, and suspect there may not be any at all}.
• I suspect that there are an infinite number of such p^2 of the form AbA, but that s strictly based on a simple heuristic argument. The next two after your
Message 2 of 7 , Aug 28, 2012
I suspect that there are an infinite number of such p^2 of the form
AbA, but that's strictly based on a simple heuristic argument.

The next two after your listed group are:

208139^2 = 43321843321

252717253^2 = 63866009963866009

Some experimentation for such numbers of the form n^2 seems to imply
that n^2 is perhaps more likely to be of the form AbA when the number
of digits in n is a multiple of 3. I can't really explain why, but I
haven't spent a lot of time thinking about it.

Statistics for the number of n with n^2 of form AbA:

n = 2 digits, solutions = 3
n = 3 digits, solutions = 4
n = 4 digits, solutions = 0
n = 5 digits, solutions = 2
n = 6 digits, solutions = 3
n = 7 digits, solutions = 1
n = 8 digits, solutions = 0
n = 9 digits, solutions = 21

On 8/28/2012 12:23 PM, woodhodgson@... wrote:
> Considering squares with an odd number of digits, the cases where the squared integers are prime and the squares less than 10^9 appear to be only these (the repetition meaning the square looks like AbA, where A is a string of digits and b any digit):
>
> 121 = 11^2, 29929 = 173^2, 69169 = 263^2, 732947329 = 27073^2.
>
> Does anyone know how uncommon these are for larger squares?
>
> {Regarding the related question about squares with an even number of digits and a complete repetition AA - I couldn't find any solutions at all for "small" squares, whether or not the original numbers were prime, and suspect there may not be any at all}.
>
>
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> ------------------------------------
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> Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
> The Prime Pages : http://primes.utm.edu/
>
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• Interesting, thank you. I am still wondering about the even AA case - I think there may be no solutions.
Message 3 of 7 , Aug 29, 2012
Interesting, thank you. I am still wondering about the even "AA" case - I think there may be no solutions.

--- In primenumbers@yahoogroups.com, Jack Brennen <jfb@...> wrote:
>
> I suspect that there are an infinite number of such p^2 of the form
> AbA, but that's strictly based on a simple heuristic argument.
>
> The next two after your listed group are:
>
> 208139^2 = 43321843321
>
> 252717253^2 = 63866009963866009
>
>
> Some experimentation for such numbers of the form n^2 seems to imply
> that n^2 is perhaps more likely to be of the form AbA when the number
> of digits in n is a multiple of 3. I can't really explain why, but I
> haven't spent a lot of time thinking about it.
>
> Statistics for the number of n with n^2 of form AbA:
>
> n = 2 digits, solutions = 3
> n = 3 digits, solutions = 4
> n = 4 digits, solutions = 0
> n = 5 digits, solutions = 2
> n = 6 digits, solutions = 3
> n = 7 digits, solutions = 1
> n = 8 digits, solutions = 0
> n = 9 digits, solutions = 21
>
>
>
> On 8/28/2012 12:23 PM, woodhodgson@... wrote:
> > Considering squares with an odd number of digits, the cases where the squared integers are prime and the squares less than 10^9 appear to be only these (the repetition meaning the square looks like AbA, where A is a string of digits and b any digit):
> >
> > 121 = 11^2, 29929 = 173^2, 69169 = 263^2, 732947329 = 27073^2.
> >
> > Does anyone know how uncommon these are for larger squares?
> >
> > {Regarding the related question about squares with an even number of digits and a complete repetition AA - I couldn't find any solutions at all for "small" squares, whether or not the original numbers were prime, and suspect there may not be any at all}.
> >
> >
> >
> >
> >
> > ------------------------------------
> >
> > Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
> > The Prime Pages : http://primes.utm.edu/
> >
> >
> >
> >
> >
> >
>
• ... obviously p^2 =concat(A,A) = A * (10^k+1) with 10^(k-1)
Message 4 of 7 , Aug 29, 2012
> Interesting, thank you. I am still wondering about the even "AA" case - I
> think there may be no solutions.
>

obviously p^2 =concat(A,A) = A * (10^k+1) with 10^(k-1) < A < 10^k
so this is not possible.

Maximilian

[Non-text portions of this message have been removed]
• True, not solvable when p is a prime. There are solutions for the general case including non-primes. I think that the smallest are (and yes, there s a
Message 5 of 7 , Aug 29, 2012
True, not solvable when p is a prime.

There are solutions for the general case including non-primes. I think
that the smallest are (and yes, there's a pattern):

36363636364^2 = 1322314049613223140496
45454545455^2 = 2066115702520661157025
54545454546^2 = 2975206611629752066116
63636363637^2 = 4049586776940495867769
72727272728^2 = 5289256198452892561984
81818181819^2 = 6694214876166942148761
90909090910^2 = 8264462810082644628100

The trick here, for these seven examples, is to find 22 digit squares
which are divisible by 100000000001 (10^11+1). The reason these are
the smallest is because 10^11+1 is the smallest number of that form
which is not square-free.

On 8/29/2012 12:31 PM, Maximilian Hasler wrote:
>> Interesting, thank you. I am still wondering about the even "AA" case - I
>> think there may be no solutions.
>>
>
> obviously p^2 =concat(A,A) = A * (10^k+1) with 10^(k-1) < A < 10^k
> so this is not possible.
>
> Maximilian
>
>
> [Non-text portions of this message have been removed]
>
>
>
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>
> Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
> The Prime Pages : http://primes.utm.edu/
>
>
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• Of course, trivial for prime p, should have seen that instantly.
Message 6 of 7 , Aug 29, 2012
Of course, trivial for prime p, should have seen that instantly.

--- In primenumbers@yahoogroups.com, Maximilian Hasler <maximilian.hasler@...> wrote:
>
> > Interesting, thank you. I am still wondering about the even "AA" case - I
> > think there may be no solutions.
> >
>
> obviously p^2 =concat(A,A) = A * (10^k+1) with 10^(k-1) < A < 10^k
> so this is not possible.
>
> Maximilian
>
>
> [Non-text portions of this message have been removed]
>
• Thanks, all clear. There is a significant gap to later candidates, as the next case where 10^k + 1 is not square-free is for k=21. The first entry in that
Message 7 of 7 , Aug 30, 2012
Thanks, all clear. There is a significant gap to later candidates, as the next case where 10^k + 1 is not square-free is for k=21.

The first entry in that family appears to be

183673469387755102041183673469387755102041 = 428571428571428571429^2

--- In primenumbers@yahoogroups.com, Jack Brennen <jfb@...> wrote:
>
> True, not solvable when p is a prime.
>
> There are solutions for the general case including non-primes. I think
> that the smallest are (and yes, there's a pattern):
>
> 36363636364^2 = 1322314049613223140496
> 45454545455^2 = 2066115702520661157025
> 54545454546^2 = 2975206611629752066116
> 63636363637^2 = 4049586776940495867769
> 72727272728^2 = 5289256198452892561984
> 81818181819^2 = 6694214876166942148761
> 90909090910^2 = 8264462810082644628100
>
> The trick here, for these seven examples, is to find 22 digit squares
> which are divisible by 100000000001 (10^11+1). The reason these are
> the smallest is because 10^11+1 is the smallest number of that form
> which is not square-free.
>
>
>
> On 8/29/2012 12:31 PM, Maximilian Hasler wrote:
> >> Interesting, thank you. I am still wondering about the even "AA" case - I
> >> think there may be no solutions.
> >>
> >
> > obviously p^2 =concat(A,A) = A * (10^k+1) with 10^(k-1) < A < 10^k
> > so this is not possible.
> >
> > Maximilian
> >
> >
> > [Non-text portions of this message have been removed]
> >
> >
> >
> > ------------------------------------
> >
> > Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
> > The Prime Pages : http://primes.utm.edu/
> >