Ponder This Challenge - March 2000 - Sum of two squares

Ponder This Challenge:

Let's borrow notation from LaTeX and let "^" denote exponentiation, so that x^2 means "x squared".

Consider the numbers 1233 = 12^2 + 33^2, and 990100 = 990^2 + 100^2.

Can you find an eight-digit number N with the same property, namely that if you break N into two four-digit numbers B and C, and add their squares, you recover N?

(N really has to have eight digits; its leading digit is not zero. B is the number formed by the first our digits of N, and C is the number formed by the last four digits of N.)

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Solution

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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 = (2B-10000)^2 +(2C-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: 17 = 4^2 + 1^2 = w^2 + x^2; 5882353 = 588^2 + 2353^2 = y^2 + z^2. Then combine them according to the usual laws (related to the multiplication of complex numbers): (w^2 + x^2)(y^2 + z^2) = (wy+xz)^2 + (wz-xy)^2 (w^2 + x^2)(y^2 + z^2) = (wy-xz)^2 + (wz+xy)^2

  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.

Solvers

• Jay Patrick Howard (3.2.2000 @ 3:51 PM EST)
• Greg Xu (3.2.2000 @ 3:51 PM EST)
• Muppala Suresh (3.2.2000 @ 4:02 PM EST)
• S. Muralidhar (3.2.2000 @ 4:46 PM EST)
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• msk (3.3.2000 @ 1:55 PM EST)
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