If (P_b)*(P_b)=31_b, where P and b represent two distinct digits 0-9 and P is one less than b, what is the value of base b?
So I have ((b-1)_b)^2=31_b
how do I do this?! please explain. :)
\(P=b-1\\ (b-1)^2 = 31_b = 3b+1\\ b^2 -2b+1 = 3b+1\\ b^2 = 5b \\ b=5 \text{ (or 0, but 0 is not a valid base)}\\ b=5, ~P=4\\ (4_5)^2 = 16_{10} = 31_5\)
