Solutions to equations Watch
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- 31-08-2016 02:04
- 31-08-2016 02:41
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- 31-08-2016 03:05
- 31-08-2016 03:23
- 31-08-2016 04:14
So going from you can work through all the possibilities of signs a and b can have and find that a must be odd and b must be even (assuming both a and b are >0 of course).
Then go back to , realising that with the LHS doesnt have an upper "cap" eg you could set a is big as you like and b = 1 and get huge numbers so we should aim to make the LHS as small as possible, therefore we should set b to be the biggest integer possible, namely
doing this and subbing into gives us;
solving this we get a= 1 and therefore b = 0 as a solution, however 0 isnt a positive integer so this solution is invalid.
Common sense says that setting b = a - 3 will not work as the LHS will only get bigger and the RHS will get smaller meaning only a negative value of b will make LHS = RHS, this negative b may not even be an integer but we dont need to worry about it, since we must have b>0 there are no valid solutions.
Edit: sat on the reply screen for an hour whilst I worked on paper, there already is an answer much better than mines, oh wellLast edited by DylanJ42; 31-08-2016 at 04:16.
- 31-08-2016 04:22
(b+k)^3 = b^3 + 3b^2 +1
The smallest number k can be is 1
in this case
(b+1)^3 = b^3 + 3b^2 + 1
b^3 +3b^2 + 3b +1 = b^3 + 3b^2 +1
3b = 0
Already the left hand side is larger than the right hand side, any larger values of k will not work for the same reason.
Therefore k is a natural number less than one - which does not exist - no k satisfies and therefore no a and b satisfy and therefore the equation has no solutions.