IBM Research | Ponder This | February 2000 solutions
Skip to main content

February 2000

<<January February March>>

February's puzzle seems to have been harder than usual. We received only one correct answer. The answer is that C must be an integer. I will assume that C is a non-integer that lies between 3 and 4. The proof generalizes but the notation gets harder. Pick an arbitrary large integer N. Consider the function f(x)=x**C, by which we denote "x raised to the C power". We will take the Taylor expansion of f(x) around the point N, and evaluate f(x) at the various points N+i as i=0,1,...,4. By hypothesis, since N+i is a positive integer, so is f(N+i). We will take a certain integer linear combination of the values f(N+i), concocted to make most of the terms of the Taylor expansion cancel out, and we will end up with a quantity which is supposed to be an integer, but instead lies strictly between 0 and 1. The Taylor expansion tells us f(N+i) = f(N) + f'(N)i + f''(N)(i**2)/2 + f'''(N)(i**3)/6 + f''''(N)(i**4)/24 + f'''''(xi)(i**5)/120, where xi is some number between N and N+i. Now multiply the integers f(N+i) by (1, -4, 6, -4, 1), respectively, and sum. So we have: 1*f(N+0) - 4*f(N+1) + 6*f(N+2) - 4*f(N+3) + 1*f(N+4). On the one hand, we are multiplying integers by integers, so our resulting sum must be an integer. On the other hand, the sum comes out as 1*f(N) - 4*(f(N) + f'(N) + f''(N)/2 + f'''(N)/6 + f''''(N)/24 + f'''''(x1)/120) + 6*(f(N) + 2 f'(N) + 4 f''(N)/2 + 8 f'''(N)/6 + 16 f''''(N)/24 + 32 f'''''(x2)/120) - 4*(f(N) + 3 f'(N) + 9 f''(N)/2 + 27 f'''(N)/6 + 81 f''''(N)/24 + 243 f'''''(x3)/120) + 1*(f(N) + 4 f'(N) + 16 f''(N)/2 + 64 f'''(N)/6 + 256 f''''(N)/24 + 1024 f'''''(x4)/120). Now, many of the terms cancel. The coefficient of f(N) is (1-4+6-4+1)=0. The coefficient of f'(N) is (0-4*1+6*2-4*3+1*4)=(-4+12-12+4)=0. Similarly the coefficients of f''(N) and f'''(N) both vanish. The f''''(N) term has a coefficient of 1: (0 - 4*1 + 6*16 - 4*81 + 1*256)/24 = 24/24 = 1. The error terms are all constant multiples of f'''''(xi). Now we know that f''''(N) = C*(C-1)*(C-2)*(C-3)*N**(C-4), and since C is between 3 and 4, we have that N**(C-4) grows smaller as N grows larger. For N large enough, f''''(N) is strictly between 0 and 1/2. (The coefficient C*(C-1)*(C-2)*(C-3) doesn't vanish because C is not an integer; this coefficient is some constant E independent of N.) The error term, involving the fifth derivative of f at various places, is bounded by some constant times N**(C-5). So the total sum is a nonzero constant times N**(C-4), plus an error term bounded by another constant times N**(C-5). For N large enough, this sum is bounded strictly away from 0, but smaller than 1 in absolute value. So we have an integer lying strictly between 0 and 1 (or between 0 and -1). By this contradiction we see that such a C cannot exist.

If you have any problems you think we might enjoy, please send them in. All replies should be sent to: