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Show that if $k$ is a natural number. And $k$ is like $4*m +3, so $k$ has a prime factor in this form. Please, show me different ways to solve this question.

Santiago
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  • Welcome to stackexchange. That said, I'm voting to close this question because it is not at all clear what the question is. If you [edit] to tell us exactly what you wish to prove and how you started and where you are stuck we may be able to help. – Ethan Bolker Jun 25 '19 at 15:27
  • If $k$ had factors of $2$, $k$ would not be $4m+3$; if $k$ had only factors of the form $4k+1$, $k$ would be $4n+1$ – J. W. Tanner Jun 25 '19 at 15:30

2 Answers2

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Hint:

The simple way uses congruences: the hypothesis means $k\equiv 3(\equiv -1)\bmod 4$, so $k$ is odd and it has only odd prime factors. Now an odd prime is congruent either to $1$ or $-1\bmod 4$.

What can you conclude as to the number of primes congruent to $-1\bmod 4$ in the decomposition of $k$ into primes?

Bernard
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"Show that if k is a natural number. And k is like 4∗m+3,so k has a prime factor in this way. Please, show me different ways to solve this question."

What question? There is no question there. Also, what do you mean by "has a prime factor in this way"? Every integer has prime factors! If m= 0, k= 3 which is a prime number. If m= 1, k= 7 which is a prime number. If m= 2, k= 9= 3*3. If m= 3, k= 15= 3*5. If m= 4, k= 19 which is a prime number.

Again, every integer has prime factors.

user247327
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  • I agree that the question is currently not stated clearly, but I surmise that what OP intended to ask for was a demonstration that $k$ has a factor of the form $4n+3$ – J. W. Tanner Jun 25 '19 at 15:43