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Re: Equal sums of terms in Arithmetic Progession
Posted:
Dec 28, 2012 6:56 PM


croomejohn@yahoo.co.uk a écrit : > 1+2+...+14 = 15+16+...+20 > > 1+4+7+10+13 = 16+19 > > 1+6+11+...+251 = 256+261+...+356 > > 1+7+13+...+55 = 61+67+73+79 > > So with d = 1,3,5 and 6 (and 10 and 16, I haven't shown these), there exists > n and m such that > > 1 + (1+d) + ... + 1+ nd = 1+(n+1)d + ... + 1+ md > > It is impossible with d = 2. Owing to 1+3+...+2n+1 = n^2 and 1+3+...2m+1 = > m^2 so m^2 = 2n^2 leads to root(2) being rational m/n.
hi,
but the right member is not "1 +..." but "a+(a+d)+(a+2d)+...+(a+md)"
so that for instance : 1+3+5+7+...+15 = 64 = 100  36 = 13 + 15 + 17 + 19
with d = 2 !
Regards.
 Philippe C. mail chephip à free.fr site divertissement mathématiques http://mathafou.free.fr



