## March 2000

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There are two different versions of the solution this month....
1.
We have N=10000B+C and N=B^2+C^2.
So 10000B+C=B^2+C^2, 10000B-B^2 = C^2-C. Also B must lie between 1000 and 9999, in order for N to be an eight-digit number. For each trial value of B in that range, compute f(B)=10000B-B^2. Let c be the square root of f(B), rounded up to the nearest integer. If our equation is satisfied, we will have f(B)=c^2-c. (The square root of f(B) will be about C-1/2, so c=C.) So we only have to do about 9000 trials to find the lone answer: B=9412, C=2353, N = 94122353 = 9412^2 + 2353^2. 2. A few readers reminded us how to solve this problem algebraically, without exhaustive search. This write-up follows S V Nagaraj's solution. Continue to let "^" denote exponentiation. The solution is obtained by considering 10000B + C = B^2 +C^2, Rewriting this as 10000^2 +1 = (2*B-10000)^2 +(2*C-1)^2, we just look at the factorization of 10000^2 + 1 = 10^8 +1 = 17 *5882353.
Express each prime factor as the sum of squares: The second equation gives an uninteresting solution. The first gives us 4705^2 + 8824^2. Since 2C-1 is odd and 2B-10000 is even, we know which is which. 2C-1=4705 so that C=2353, and 2B-10000=8824 (or -8824) so that B is 9412 (or 588). B=588 is ruled out by demanding a strict 8-digit number. So the answer is B=9412, C=2353, N=94122353. |

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