My teacher has given my class a challenge task: to prove the following implication:
Let and be integers such that and .
Prove: If is prime, then there exists an integer such that
the following identity was given to us to use:
Let and be integers and , then
<------ note, there should be a on the top of the
This is for odd but not for even. Why? Because if is even, and you go then your last term is going to be not In that case multiplying by will not give . Only when is odd does the result stand.
I didnít mention this because I thought you might want to check everything carefully yourself and make sure that what I wrote made sense.
I also wrote It might look okay, but donít be fooled. It might still break down if certain conditions are not met Ė for example, what if Luckily the question says so and weíre safe. Everything has to be carefully checked and every gap must be covered.