Prove that every even number $k\geqslant 8$ can be represented as $m,n\in \mathbb{N}\setminus \{1\}$ $$k=m+n$$ and $\gcd(m,n)=1$.
I was able to do it, if $k$ is an odd number but not if it is even. I could use some hints.
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1Hint: if $m$ and $n$ are close to each other, they can't have many factors in common... – Théophile Feb 17 '15 at 21:33
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Another hint: Try some examples first. How can you write 12 as a sum of two relatively prime numbers? What about 10? – mweiss Feb 17 '15 at 21:37
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Further hint: $,m,n,$ have equal parity so, being coprime, both are odd. So consider close odds. – Bill Dubuque Feb 17 '15 at 21:37
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If $k = 4m: m \in \Bbb{N}$ then $k = (2m-1) + (2m+1)$ and both of those terms are odd and differ by exactly $2$, so one step of Euclid's algorithm shows that $$\gcd(2m+1,2,-1) = \gcd(2m-1,2) = 1.$$
Having taken care of all multiples of $4$, now consider $k = 4m+2$. In that case, $k = (2m-3) + (2m+1)$ and $$\gcd(2m+1,2,2m-3) = \gcd(2m-3,4) = 1$$ since $2m-3$ contains no powers of $2$.
Mark Fischler
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I changed the four instance of \text{gcd} to \gcd in this answer. The latter provides proper spacing between $\gcd$ and whatever precedes it. Contrast $a\text{gcd}(b,c)$ with $a\gcd(b,c)$. They look different. ${}\qquad{}$ – Michael Hardy Feb 17 '15 at 21:40
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Hint: If $\mathrm{gcd}(k, m) = 1$, then if $q$ is an integer such that $q \mid m$ and $q \mid k-m$, then $$q \mid (k-m) + m = k$$
Can you deduce that $m$ and $k-m$ are coprime?
Mathmo123
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This is the germ of the idea behind Stieltjes generalization of Euclid's proof that there are infinitely many primes, and it often proves useful. – Bill Dubuque Feb 17 '15 at 21:58
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Assume $k\geq 8$. By Bertrand's postulate (http://mathworld.wolfram.com/BertrandsPostulate.html), there exists a prime number $p$ between $\frac{k}{2}$ and $k-2$. Now $p$ and $k-p$ clearly are coprime.
Uncountable
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