Sum of Two SquaresDate: 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 Check out our web site! http://mathforum.org/dr.math/ |
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