Submitted by maths123 on Tue, 05/19/2015 - 18:08
Forums:
In the second part of the question, I reached the below stage:
$ kz = z^4 - 3z^2 x + 3x^2 $
The mark scheme says that $ z^3 > k $ - how do I go about proving this?
My main attempts so far have been trying to replace the two variables with $ x $ or $y$ and using the fact that these are positive integers.
Also I am not sure when to use the equation $ x+y = z^2 $
Hints
You have the right idea in using the fact x and y are strictly greater than zero and what you have is basically the answer. Working from
$$ z^3 = k^2 - 3kx +3x^2 $$
it's probably more useful to use the substitution $ k = x + y $ in this and keep the term in $k^2$ you want for the inequality. From there should find
$$ z^3 = k^2 - ("positive-stuff") $$
which gives you the answer. Your best bet in these questions where you are given the answer is to try and meddle with the stuff that isn't in the result (here $-3kx +3x^2$) and leave terms you're aiming to end up with (a $z^3$ and a $k^2$) alone.
Replicating the first part
To make the second part work, one method is to imitate the first part as far as possible. In the first part we tried to find a perfect square. We want this perfect square to account for all the x terms, just like in the first part, so can you fiddle around with adding some things and multiplying by constants to express the right hand side of the expression $$kz=z^4-3z^2+3x^2$$ that you've derived to write it as a square? Once you have, it should hopefully be clear how you get that $z^3>k$.