# Decimal and nondecimal almost perfect squares of the form aaabbb

by K Sengupta
Tags: aaabbb, decimal, form, nondecimal, perfect, squares
 P: 108 Let the perfect square = $$N^2$$. We have $$N^2$$ = aaabbb + 1 (N+1)(N-1) = 111 X a00b = 111(999a+a+b) If 9a is a perfect square, say $$c^2$$, 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.