Guest Posted March 11, 2010 Report Share Posted March 11, 2010 (edited) Determine all possible triplet(s) (p, q, r) of positive integers, with p < q < r, that satisfy the equation: p2+ q2= r2 - 23, where q is the harmonic mean of p and r. Edited March 11, 2010 by K Sengupta Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 11, 2010 Report Share Posted March 11, 2010 (edited) From p < q < r and p2 + q2= r2 - 23 => p2 + q2 - r2 = - 23 we understand that it is impossible the equation to be true even if p, q and r are negative. And they must be all positive integers. IMPOSSIBLE! Edited March 11, 2010 by Ianis G. Vasilev Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted March 11, 2010 Report Share Posted March 11, 2010 From p < q < r and p2 + q2= r2 - 23 => p2 + q2 - r2 = - 23 we understand that it is impossible the equation to be true even if p, q and r are negative. And they must be all positive integers. IMPOSSIBLE! (p,q,r)=(2,3,6)? That works. So, it's not impossible! Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 12, 2010 Report Share Posted March 12, 2010 Whoops! I was wrong as I wrote down on my piece of paper that p > q > r. Sorry. Quote Link to comment Share on other sites More sharing options...
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Determine all possible triplet(s) (p, q, r) of positive integers, with p < q < r, that satisfy the equation: p2+ q2= r2 - 23, where q is the harmonic mean of p and r.
Edited by K SenguptaLink to comment
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