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wu :: forums    riddles    easy (Moderators: towr, Eigenray, ThudnBlunder, Icarus, Grimbal, SMQ, william wu)    Prime Works « Previous topic | Next topic » Pages: 1  Reply Notify of replies Send Topic Print    Author  Topic: Prime Works  (Read 279 times) The_Fool Guest Prime Works   « on: Mar 8th, 2004, 10:06am » Quote Modify Remove is it possible for (p^2+q^2)/(pq) to be an integer when p does not equal q?  When p=q, it's 2.  What about when their not? IP Logged Sir Col Uberpuzzler impudens simia et macrologus profundus fabulae     Gender: Posts: 1825 Re: Prime Works   « Reply #1 on: Mar 8th, 2004, 10:19am » Quote Modify Assuming you meant p,q are prime...   :: If (p2+q2)/(pq)=r, where r is integer, we get p2+q2=pqr.   As p divides RHS, p must divide LHS. However, q and p are different, so p cannot divide q. Hence there is no solution. ::   [e] I thought this was familiar and did a quick search. It seems that I started a thread, asking the more general question about the case where p and q are not necessarily prime... http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_eas y;action=display;num=1073053801 [/e] « Last Edit: Mar 8th, 2004, 10:25am by Sir Col » IP Logged mathschallenge.net / projecteuler.net Pages: 1  Reply Notify of replies Send Topic Print Forum Jump: ----------------------------- riddles -----------------------------=> easy   - medium   - hard   - what am i   - what happened   - microsoft   - cs   - putnam exam (pure math)   - suggestions, help, and FAQ   - general problem-solving / chatting / whatever ----------------------------- general -----------------------------  - guestbook   - truth   - complex analysis   - wanted   - psychology   - chinese « Previous topic | Next topic »

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