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# Solutions to equations Watch

1. The question asks to prove that there are no solutions to the equation if and are both positive integers.
Any ideas?
2. (Original post by Ano123)
The question asks to prove that there are no solutions to the equation if and are both positive integers.
Any ideas?
This can be rewritten as a^3 = (b+1)^3 - 3b -> (b+1)^3 - a^3 = 3b -> (b+1-a)(b^2+2b+1+ab+a+a^2) = 3b. Maybe you can do something with that? I really have no idea.
3. (Original post by Ano123)
The question asks to prove that there are no solutions to the equation if and are both positive integers.
Any ideas?
If then obviously it doesn't hold as you would get thus no real solutions.

Say - then so it doesn't hold as the LHS is negative while RHS is positive.

If - then this becomes slightly trickier for a brain of mine at 2am. Try and figure it out from there.
4. Clearly (real) as the RHS is obviously greater than . We find values of b such that .
If . Therefore . So if then . Of course no closed interval of length 1 contains three distinct integers. So any solutions if they do exist will have
This is very similar to a STEP question,
5. (Original post by Ano123)
The question asks to prove that there are no solutions to the equation if and are both positive integers.
Any ideas?
I have a super ugly proof I think

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 well
6. (Original post by Ano123)
The question asks to prove that there are no solutions to the equation if and are both positive integers.
Any ideas?
let a = b+k (positive integers so a is clearly larger than b and k is a natural number greater than 0)
(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.

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Updated: August 31, 2016
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