Diophantine equations in Number TheoryDate: 01/24/2001 at 14:51:39 From: Kristi Harp Subject: Diophantine equations in number theory Here is the problem: If a and b are relatively prime positive integers, prove that the Diophantine equation ax-by = c has infinitely many solutions in the positive integers. [Hint: There exist integers x0 and y0 such that ax0+by0 = c. For any integer t, which is larger that both |x0|/b and |y0|/a, a positive solution of the given equation is x = x0+bt, y = -(y0-at).] (Note: x0 is x null or naught...I didn't know how else to write it.) I'm stuck on this homework problem. If anyone can please help me I would greatly appreciate it. Thank you. Date: 01/24/2001 at 19:35:01 From: Doctor Anthony Subject: Re: Diophantine equations in number theory If a and b are coprime you can always find integers p, q such that ap + bq = 1 using Euclid's algorithm. Then multiply by c to get ax' + by' = c ......(1) where x' = cp, y' = cq and finally let x = x'+bt y = -(y'-at) and substituting into the equation ax - by = c we get a(x'+bt) +b(y'-at) = c ax' + abt + by' - abt = c ax' + by' = c which we have shown from (1) is true. It follows that there are an infinity of solutions of the form x = x'+bt y = -(y'-at) where ax'+by'=c - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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