I started with an even odd arguement, where I tried to show that no matter which numbers you pick, the sum would always be even. However, this doesn't work for the arrangment of 3 evens and one odd. Where do I go from here?
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Write $p=a+b+c+d$ and say $p$ is prime. Then we have $$ab=c(p-a-b-c)$$ so $$(a+c)(b+c) = cp$$ which means that $$p\mid a+c\;\;\;\;{\rm or}\;\;\;\;p\mid b+c$$ in 1st case we get $p=a+b+c+d\leq a+c$ a contradiction. The same contradiction we get in the second case. So $p$ must be composite.
nonuser
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Where did you get $ab=c(p-a-b-c)$ from? And how does that leaed to $(a+c)(b+c)=cp$? – Gerard L. Feb 09 '18 at 22:32
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$p=a+b+c+d$ so $d= p-a-b-c$... – nonuser Feb 09 '18 at 22:33
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@GerardL. as for the second half of your question, multiply out $c(p-a-b-c)$ and then move the last three terms to the LHS. – Steven Stadnicki Feb 09 '18 at 22:37