Q: Let b, c, and f be intergers. If gcd(b,c) = f, then there are integers t and
ID: 2940775 • Letter: Q
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Q: Let b, c, and f be intergers. If gcd(b,c) = f, then there are integers t and w such that bc = b^2 t + c^2 w (notice that the conclusion does not involve f at all, we just gave the gcd a name to make it easier to start the proof). Here's how I started it: Assume gcd(b,c) = f Then by The GCD Theorem, there are integers n, m, s, and r such that b=fn c=fm f=bs+cr Thus, bc = (fn)(fm) = (bs+cr)n * (bs+cr)m = (bsn+crn) * (bsm+crm) = b^2 s^2 nm + bsncrm + crnbsm + c^2 r^2 nm = b^2 (s^2 nm) + 2 bsncrm + c^2 (r^2 nm) But I couldn't get rid of (2 bsncrm) in order to make it like b^2(some integer) + c^2(some integer) Thanks a lot in advance! Q: Let b, c, and f be intergers. If gcd(b,c) = f, then there are integers t and w such that bc = b^2 t + c^2 w (notice that the conclusion does not involve f at all, we just gave the gcd a name to make it easier to start the proof). Here's how I started it: Assume gcd(b,c) = f Then by The GCD Theorem, there are integers n, m, s, and r such that b=fn c=fm f=bs+cr Thus, bc = (fn)(fm) = (bs+cr)n * (bs+cr)m = (bsn+crn) * (bsm+crm) = b^2 s^2 nm + bsncrm + crnbsm + c^2 r^2 nm = b^2 (s^2 nm) + 2 bsncrm + c^2 (r^2 nm) But I couldn't get rid of (2 bsncrm) in order to make it like b^2(some integer) + c^2(some integer) Thanks a lot in advance!Explanation / Answer
f=bs+cr Thus, bc = (fn)(fm) = (bs+cr)n * (bs+cr)m = (bsn+crn) * (bsm+crm) = b^2 s^2 nm + bsncrm + crnbsm + c^2 r^2 nm = b^2 (s^2 nm) + 2 bsncrm + c^2 (r^2 nm) dear friend you are on correct path.... now take these steps.. bc - 2bcsnrm = b^2 (s^2 nm) + c^2 (r^2 nm) bc (1-2snrm) = b^2 (s^2 nm) + c^2 (r^2 nm) bc = b^2 (s^2 nm / (1-2snrm)) + c^2 (r^2 nm / (1-2snrm)) bc = b^2 t + c^2 w hope you got answer..
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