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Sum of Two Squares


Date: 12/04/97 at 00:07:46
From: Mary Writz
Subject: Sum of two squares

Here is a problem we have been working on in my math class.

The number 65 can be expressed as the sum of two squares in two 
different ways: 65 = 8'2 + 1'2 = 7'2 + 4'2.  

What is the smallest number that can be expressed in twelve different 
ways as the sum of two squares?

Any help would be greatly appreciated.


Date: 12/04/97 at 10:11:32
From: Doctor Rob
Subject: Re: Sum of two squares

I assume you do not consider 8^2 + 1^2 = 65 = 1^2 + 8^2 different 
ways, but see below if this assumption is false.

I assume you mean the squares of nonnegative numbers, else you would 
say that 65 can be expressed as the sum of two squares in eight 
different ways. Again, see below if this assumption is false.

If a number 2*n has k > 0 representations in this form, then n will 
also have k representations, and be smaller, so we can eliminate all 
even n's. Likewise, if p*n has k > 0 representations in this form, and 
p is of the form 4*j+3, then p|n, and N = n/p = (p*n)/p^2 will also 
have k representations, and be smaller, so we can eliminate all n's 
with such a prime factor. Thus our attention is focused on n's all of 
whose prime divisors have the form 4*j+1.

Now any such n which has 24 divisors would have this property. If the
factorization looks like Product(p(i)^(e(i))) for i = 1 to s, then the
number of divisors will be Product(e(i)+1) for i = 1 to s.  For this 
to equal 24, there are seven possibilities:

Factors   s   e(1)  e(2)  e(3)  e(4)
  24      1    23
 12*2     2    11     1
  8*3     2     7     2
  6*4     2     5     3
 6*2*2    3     5     1     1
 4*3*2    3     3     2     1
3*2*2*2   4     2     1     1     1

Clearly using the smallest primes with the largest exponents is best, 
so p(1) < p(2) < p(3) < p(4). The four smallest primes congruent to 
1 mod 4 are 5, 13, 17 and 29.  Thus our candidate numbers are:

   a = 5^23,
   b = 5^11*13,
   c = 5^7*13^2,
   d = 5^5*13^3,
   e = 5^5*13*17,
   f = 5^3*13^2*17,
   g = 5^2*13*17*29.

It turns out that g < f < e < d < c < b < a, so apparently g is the 
answer. It is certainly true that g has 12 different expressions as 
the sum of two squares of nonnegative numbers:

   160225 = 400^2 + 15^2,
          = 399^2 + 32^2,
          = 393^2 + 76^2,
          = 392^2 + 81^2,
          = 384^2 + 113^2,
          = 375^2 + 140^2,
          = 360^2 + 175^2,
          = 356^2 + 183^2,
          = 337^2 + 216^2,
          = 329^2 + 228^2,
          = 311^2 + 252^2,
          = 300^2 + 265^2.

If you *do* consider 8^2 + 1^2 = 65 = 1^2 + 8^2 to be different, then 
you want such an n with 12 divisors, and the smallest one seems to be 
5525 = 5^2*13*17, with (2+1)*(1+1)*(1+1) = 3*2*2 = 12 divisors.

   5525 = 74^2 + 7^2,
        = 73^2 + 14^2,
        = 71^2 + 22^2,
        = 70^2 + 25^2,
        = 62^2 + 41^2,
        = 55^2 + 50^2,
        = 50^2 + 55^2,
        = 41^2 + 62^2,
        = 25^2 + 70^2,
        = 22^2 + 71^2,
        = 14^2 + 73^2,
        = 7^2 + 74^2.

If order doesn't matter, but signs do, then the answer is
325 = 5^2*13,

   325 = 18^2 + 1^2,
       = 18^2 + (-1)^2,
       = 17^2 + 6^2,
       = 17^2 + (-6)^2,
       = 15^2 + 10^2,
       = 15^2 + (-10)^2,
       = (-15)^2 + 10^2,
       = (-15)^2 + (-10)^2,
       = (-17)^2 + 6^2,
       = (-17)^2 + (-6)^2,
       = (-18)^2 + 1^2,
       = (-18)^2 + (-1)^2.

If both order and signs matter, then the answer is 25 = 5^2.

   25 = 5^2 + 0^2,
      = 4^2 + 3^2,
      = 4^2 + (-3)^2,
      = 3^2 + 4^2,
      = 3^2 + (-4)^2,
      = 0^2 + 5^2,
      = 0^2 + (-5)^2,
      = (-3)^2 + 4^2,
      = (-3)^2 + (-4)^2,
      = (-4)^2 + 3^2,
      = (-4)^2 + (-3)^2,
      = (-5)^2 + 0^2.

-Doctor Rob, The Math Forum
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Associated Topics:
High School Number Theory

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