Can you solve the following equation?
½ = 1/x² + 1/y² +...+ 1/z²
All variables must be different, positive integers, and there must be a finite number of terms.
(In reply to
Less than 9? by e.g.)
You can't do it with 6 numbers either.
Call them a,b,c,d,e,f where a<b<c<d<e<f
As seen before, a has to be 2.
We can also see that b has to be 3, because if we take 2,4,5,6,7,8 then we get a number less than 1/2.
By the same argument, c has to be 4, d has to be 5, and e has to be 6. So that leaves us only to determine f.
Plugging the above numbers in, we get the sum for a through e is 0.4913888..., which doesn't leave room for f to be any whole number.
By a similar argument, if we assume there are 7 numbers: a,b,c,d,e,f and g, we know that a=2, b=3, c=4, and d=5.
Edited on November 9, 2004, 10:50 am

Posted by Avin
on 20041109 10:46:57 