Decimal and nondecimal almost perfect squares of the form aaabbb


by K Sengupta
Tags: aaabbb, decimal, form, nondecimal, perfect, squares
K Sengupta
K Sengupta is offline
#1
Nov20-08, 02:28 AM
P: 110
Determine all possible positive decimal integer(s) of the form aaabbb, each with no leading zeroes, that becomes a perfect square when 1 is added to it.

What are the positive nondecimal integer base(s) S, with S<=16, such that S admits at least one valid solution in conformity with the given conditions?

Note: a cannot be equal to b.
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wywong
wywong is offline
#2
Mar3-09, 04:03 AM
P: 107
Let the perfect square = [tex]N^2[/tex]. We have

[tex]N^2[/tex] = aaabbb + 1
(N+1)(N-1) = 111 X a00b
= 111(999a+a+b)

If 9a is a perfect square, say [tex]c^2[/tex], then by choosing N=ccc+1 and b=2c-a, the above equation will be satisfied.

Since aaabbb must be in the range 111000 to 999888, N must be in the range 334 to 999. Thus a can only assume values 1 and 4, with c = 3 and 6 respectively. Indeed, 111555 and 444888 are solutions.

However, N may not necessarily be of the form ccc+1. Although we only need to consider the cases when (N-1) is divisible by 37 and cases when (N+1) is divisible by 37, we still have 30+ cases to consider.

I can't find a way to substantially reduce the number of cases to consider, so I wrote a simple computer program to try them out. The solutions are:

base 5: aaabbb = 111333, N = 223
base 9: aaabbb = 222666, N = 445
base 10: aaabbb = 111555, N = 334; aaabbb = 444888, N=667
base 13: aaabbb = 333999, N = 667
base 16: aaabbb = 555888, N = 93D

As seen in the case of base 16, N is not of the form ccc+1. So it seems that many separate cases have to be considered.


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