Proving Perfect SquaresDate: 07/05/98 at 16:53:34 From: Gary Brisebois Subject: Algebra Suppose a, b, and c are positive integers, with no factor in common, where 1/a + 1/b = 1/c. Prove that a + b, a - c, and b - c are all perfect squares. I have spent the last three days of my spare time trying to solve this one. Maybe it has been too long since my university days... The most promising route I devised was to consider each integer as the product of primes, where no prime in one could be found in both of the others: Since a + b = ab/c it follows that a + b = P1^(x1+y1-z1) * ...Pn^(xn+yn+zn) where P1 thru Pn is the set of all primes found as factors of a, b, or c, and where xm, ym, and zm are the exponents of each of those primes found in each of the integers respectively. It remains to show that for each m, xm+ym-zm is even. This question was published in a local paper as an example of questions asked of Canada's best high school students. I am stumped; please help! Date: 07/07/98 at 12:24:27 From: Doctor Wilkinson Subject: Re: Algebra This is a pretty tricky problem. You have the right general idea: you want to show, for example, that in the prime factorization of a + b, every prime appears to be an even power. Let's write our equation in the form (a+b)c = ab Note that if a prime p divides a+b, then it divides the left side of the equation so it must divide the right side also, so it must divide ab and therefore must divide either a or b. But if it divides a and a+b then it divides b, and similarly if it divides b and a+b then it divides a, so it must divide a and b in any case. Now let a = p^k a' and b = b^l b', where p does not divide either a' or b'. Now I'm going to let you take over. Substitute back into the original equation and try to show that k = l and therefore that p^2k is the highest power of p dividing a + b. The other two parts of the problem are exactly similar. Let me know if you get stuck. - Doctor Wilkinson, The Math Forum http://mathforum.org/dr.math/ |
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