2005 base 10 is not a square. Neither is 2005 base 7 a square (equal 2*7^3+5=691). Is there any base b such that 2005 base b is a square?
Let 2x^3 + 5 = y^2
Clearly, 2x^3+5 is odd. Then, y^2 is odd, and therefore is the square o f an odd number.
Thus, 2x^3+ 5 = (2z+1)^2
=> 2x^3 = 4z^2 +4z -4
=> x^3 = 2(z^2+z-1)
Hence, x must be even. Therefore, the rhs is an even number divisible by 8.
However, z^2+z =z(z+1). Therefore, z^2+z-1 must be an odd number. Therefore, 2(z^2+z-1) is NOT divisible by 8.
This is a contradiction.
Consequently, there does NOT exist any base b, such that 2005_ base b is a perfect square.
Edited on September 11, 2023, 7:52 am
Puzzle Solution
