Matematické Fórum

Skrytý text:
↑ stuart clark:

Let be an integer such that . Then



Since is a prime then or .

1. Let . Then and where is an integer. If we put this identity into the fist equation then we obtain that



a) If then the expression is a composite number. Therefore and must be true. This implies that



There are two solutions of this inequality, if we consider that , i.e. . Let us substitute each value of into (*):



Thus there is a solution .



and this equation has no solution in the set of all primes.

b) If then . Similarly to the previous case, let us set each of these values into (*):



Thus there is another solution .

- no solution

- no solution

- no solution

.

2. If we consider that we can show in the same way that there exists one more solution .

Skrytý text:

1. If we wanted to classify the topic, then the number theory would be the correct sector.

2. Any integer square can be looked for as for an integer , so why not to require for suitable integer (and similarly in all other rings).

      ... So it is not a consequence of .

      ... I assume your thoughts originated from a fact that and must "appear somehow", therefore -- that argument can be avoided (anyhow, one can doubt of that, and something more than "it is obvious" should be shown).

3. An introduction of shifts the question from natural numbers to integers. So your discussion is not complete. can mean also , , . It is easy to see that k must be positive. But at least a case that is impossible should be mentioned explicitely, I guess.