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Then if 2uv = 2u'v' and u 2 - v 2 = U,2 - v,2, it follows that u = u'. Hence primitive Pythagorean triangles are generated from the formulas without duplication. 7. DIOPHANTINE EQUATIONS 25 was proved first by 1. ) 4. Prove that the area of any Pythagorean triangle is divisible by 6. 5. How many Pythagorean triangles are there with hypotenuse less than 120 ? 6. Find all positive integers a and b such that a2 + b2 = 65 2 • 7. Where a, b, and c are natural numbers, a2 + 2b2 = c2 and (a,b,c) = 1 iff for some natural numbers u and v with (u,2v) = 1, 8.
Since a,2 has the form 4d or 4d + 1, it follows that a12 + b2n has the form 4e + 2 or 4e + 3. But (V/2)2 does not have either of these forms. Contradiction. Hence a is odd. Let D = 2abv'k. Since k is a square integer, D is an integer. Let A = k/ D, B = 8a 2b2n/ D and C = (a 4 +16b4 n 2)/ D. Then A2+B2 = C2 and ~AB = n. Since a is odd, so is a4 + 16b4 n 2. Thus, since (a, b) = 1, we have (2b, a4 + 16b4 n 2) = 1. Thus the numerator of C (when it is expressed as a fraction in lowest terms) is at least a4 + 16b4 n 2 av'k and this is greater than a, the numerator of the original hypotenuse, as a straightforward calculation reveals.
FERMAT'S LAST THEOREM + A(pb ± aq)2 = (p2 + Aq2)(a 2 + Ab2) (2) If the prime p2 + Aq2 divides a 2 +Ab2, (1) implies that it divides pb± aq (pa =f Aqb)2 for one of the signs, and (2) then implies that it divides pa =f Aqb for the corresponding sign. 2 Let x and y be positive integers such that xy has the form a 2 +3b2 but x does not. If x is odd then y has an odd prime factor not of that form. e. they are both even or both odd), and a 2 + 3b2 has the form 4c, that is, xy is divisible by 4. If a and b are even then xy/4 = (a/2)2 + 3(b/2)2 has the original form.