MATH324 Midterm: MATH 324 UofA Midterm 2 Solution
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Show all your work: continue on the back of the page if you run out of space. Solution: write n = 3ah where 3 6 |h and a 0. Therefore, (3n) = 3 (n) if and only if 3|n: (20 points) find all the solutions to where x, z are positive integers and (x, z) = 1 x2 + 144 = z2. Solution: this is a simpler version of problem 13. 1. 11 from homework. The problem is asking for pythagorean triples with y = 12. In fact, the condition that (x, z) = 1 implies that we are only looking for primitive pythagorean triples. Then there are 0 < k < l integers, (k, l) = 1 and k 6 l(mod 2) such that x = l2 k2, y = 2kl, z = k2 + l2. Because y = 12 = 2kl, we know that kl = 6.
