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Given any non-zero $a,b$ and $n$, how to show that there exist integer $k$ and $l$, such that $gcd(k,l)=1$ and $n\mid ak+bl$.

Any help is highly appreciated.

Thanks in advance.

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    Given that $a,b,n$ are arbitrary, $k$ and $l$ should be chosen such that $a\vert k$ and $a\vert l$; then $a$ divides any linear combination of $k,l$. – 2017-02-19
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    @Akay In that case $gcd(k,l) \neq 1.$ – 2017-02-19
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    Oh..yes. What if we try cases: If at least one of $a,b$ is divisible by $n$, then at least one of $k,l$ is 1, (suppose $n|a$, so $k=1$)and then $l$ can be chosen s.t. $n|bl$. All that is left is to show existence in case $n$ divides neither of $a, b$. – 2017-02-19
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    @Akay The most important case is the last one that you mentioned. – 2017-02-19
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    @bor: Are you sure the claim is true? – 2017-02-19
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    "All that is left to show is..." That has to be one of the most blithe understatement I've read today. – 2017-02-19

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Let $d=gcd(a, b)$.

Let $k=b/d$ and $\ell = (bn-1)a/d$. Obviously $n \mid ka + \ell b$ and $gcd(k, \ell) = 1$.