I've worked out the reasoning, but how do I write the proof?

Note that $${b\over 4}+{1\over a}={a\over 20}\implies a^2=5ab+20\implies a|20 , 5|a$$so all the possible cases are $$a\in \{-20,-10,-5,5,10,20\}$$by a simple investigation, we conclude there are no integers $a,b$ such that $${b\over 4}+{1\over a}={a\over 20}$$

P.S.

Generally, all the answers to the equation $${a\over 4}+{1\over b}={c\over 20}$$are as follows$$(a,b,c)=\left(a,b,5a+{20\over b}\right)$$with any arbitrary $a\in \Bbb Z$ and $b|20$.

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Proof Writing

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