[seqfan] a somewhat silly problem
Jeffrey Shallit
shallit at uwaterloo.ca
Fri Aug 21 15:45:37 CEST 2020
Here's a problem that occurred to me (thinking about programming
problems I could assign in a course).
2^8 = 256, and the base-10 digits 2,5,6 can be permuted
to form 625 = 25^2.
What other powers of 2, 2^n, have the property that one can rearrange
the decimal representation to form a square (other than 2^{n/2}
squared)? I don't allow leading zeros.
It is easy to see, by comparing mod 9, that n must be even.
Here are the ones I've found so far:
2^8, 25^2
2^10, 49^2
2^12, 98^2
2^14, 178^2
2^20, 1028^2
2^26, 8291^2
2^28, 19112^2
2^30, 33472^2
2^32, 51473^2
2^34, 105583^2
2^36, 129914^2
2^38, 640132^2
2^40, 1081319^2
2^42, 1007243^2
2^44, 3187271^2
2^46, 4058042^2
2^48, 10285408^2
2^50, 32039417^2
2^52, 44795066^2
2^54, 100241288^2
This suggests that perhaps every sufficiently large even exponent will
work. Can anyone prove that?
I submitted the sequence of exponents as A337252 in the OEIS.
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