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I Don't Even Know How to Title This...
We here at Team Today have mentioned before that Justin Freeman is the brains of the team. Here's the proof. No pun intended. Basically looks like a jazzed-up Pythagorean Theorem to us...
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\r\nHow do I pass the time between races on the World Cup? Sometimes I prove theorems from number theory. Here is a proof of Fermat’s Two Square Theorem I came up with on my first World Cup trip in 2002.
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\r\n(Yes, I know that unlike his more famous Last Theorem, Fermat proved this one himself centuries ago).
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\r\n(Yes I am 100% serious about all of this).
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\r\nTheorem: Every prime of the form p=4k+1 (k an integer) can be expressed as the sum of two unique perfect squares.
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\r\nProof:
\r\nFirst we need to consider the group U(p). [In this first part I am duplicating a result of Gauss that showed that is n divides the order of U(p) then there are (x) elements in U(p) of order n.]
\r\nU(p) has order 4k so we know that by the Fundamental Theorem of Finite Abelian Groups, U(p) ≈ Z(2n) (+) G0, where n>2 and G0 has odd order or U(p) ≈ Z(2a) (+)Z(2b)(+) G1.
\r\nIn the second case, U(p) would have more than one element of order 2. This implies that x2=np+1 has more than two solutions, thus (x+1)(x-1)=np has more than one solution. For each of these solutions, p divides x+1 or p divides x-1 and 0\r\n
\r\nNow define q to be the greatest integer less than the square root of p. Our goal now is to find a number x, 0\r\n
\r\nTo show that such an x exists, consider the set {s,2s,3s,…,qs}. If there exists i,j ≤q, such that is-js=r mod p, r≤q, then (i-j)s=r and (j-i)s*=r, so we can define x=│i-j│, y=r. If such an i and j exist, we are done. So assume they do not.
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\r\nThen rewrite the set {s,2s,3s,…qs} mod p as {y1,y2,y3,…,yq} where yj>yi when j>i and yqq+1, yi+1>yi+q+1, so that yq≥q2+q.
\r\nHowever, it also must be true that yq\r\nThus in order for there not to be a solution we must have q2+q≤yq≤p-q, which can be rewritten as p>q2+2q or, since p and q are integers, q2+2q+1≤p, or (q+1)2≤p. This contradicts the definition of q, showing a contradiction of our assumption that there is no solution, so there must indeed exist x,y such that x2+y2=p.
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\r\nTo show that this solution is unique, assume that x2+y2=p and x*2+y*2=p. We know that y=sx mod p and y*=sx* mod p. So (y-y*) = s(x-x*) mod p which means (y-y*)2+(x-x*)2=0 mod p. Since the difference between y and y* is less than q, as is the difference between x and x*, this means (x-x*)2+(y-y*)2=p=x2+y2+x*2+y*2-2(xx*+yy*). So 2(xx*+yy*)=p, which means 2 divides p, which is a contradiction.
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\r\nEd.'s Note: We apologize if some of the symbols and thingys(?) didn't come out correctly. If you want the original copy of this proof, I'll email it to you. Contact me at bode1978@yahoo.com
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Written By: JFreemanDate Posted: 5/18/2004Number of Views: 343
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