Let be the set of all positive integers such is if, then it implies that there exists positive integers such as,Originally Posted by DMT
. Your aim is to show that this set is empty, i.e. the diophantine equation got no solution. Assume it is not then by the well-ordering principle there exists a minimal element in call it . Then by definition of we have,
Note that the right hand side is divisible by 3. That means one of two things.
1)Both divisible by 3.
2)The integers leave remainders 1 and 2
If case 1) is true then we have,
Now the left hand side is divisible by 3, thus because 3 is prime. Thus,
with which is impossible because of the minimal nature of . (this reminds me of Fermat's infinite descent).
From this discussion we proved that condition 1) cannot be true.
If case 2) is true then we have, for some integers that,
Open parantheses and organize,
Important note that the left hand side is divisible by 3, thus, thus, because 3 is prime. As a result we have that 3 divides . This is your problem.
I know I did not answer your question. I wrote it so that the next time I work on this problem I can continue from here.