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Re: Equal sums of terms in Arithmetic Progession
Posted:
Dec 28, 2012 7:34 PM


On Sat, 29 Dec 2012 00:56:24 +0100, Philippe 92 wrote: > 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.
> 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 !
True, although that example doesn't start out on the right side with the next term after the last one on the left side. The algebra for m^2 = 2n^2 is that LHS = n^2, RHS = m^2  n^2, LHS = RHS, so m^2 = 2n^2.
 jiw



