Theorem: If $a|m$ and $a|n$ then $a|mx+ny$, where $a,m,n,x,y\in \mathbb{Z}$.
I was studying this result from number theory recently. I was wondering whether the inverse and converse and contrapositive of this theorem is also true. I have seen some examples where it seemed to me that these theorems are being used but there was nothing mentioned except for the above theorem. I am not sure. I don't remember exactly but it was something along the lines of $p_i\nmid p_1p_2\cdots p_k+1$ where $i\in\{1,2\cdots, k\}$.
Inverse of the theorem: If $a\nmid m$ and $a\nmid n$ then $a\nmid mx+ny$ , where $a,m,n,x,y\in \mathbb{Z}$.
Converse of the theorem: If $a|mx+ny$ then $a|m$ and $a|n$, where $a,m,n,x,y\in \mathbb{Z}$.
Contrapositive of the theorem: Let $a,m,n\in\mathbb{Z}$. If there exist $x,y\in \mathbb{Z}$ such that $a\nmid mx+ny$ then $a\nmid m$ and $a\nmid n$.
Is the inverse, converse and contrapositive of this theorem true?
Your statement, with variables restricted to Z, is
a|m and a|n implies a|mx + my.
The inverse is
not(a|m and a|n) implies not(a|mx + my)
which is equivalent to
(not-a|m or not-a|n) implies not(a|mx + my).
The contrapositive is
not(a|mx + my) implies not(a|m and a|n)
which is equivalent to
not(a|mx + my) implies (not-a|m or not-a|n).
Were the statement considered to be
a|m and a|n implies for all x,y, a|mx + my
then inverse is
not(a|m and a|n) implies not(for all x,y, a|mx + my)
which is equivalent to
(not-a|m or not-a|n) implies exists x,y with not-(a|mx + my),
The converse is
(for all x,y, a|mx + my) implies a|m and a|n,
and the contrapositive is
not(for all x,y, a|mx + my) implies not(a|m and a|n)
which is equivalent to
exists x,y with a|mx + my) implies (not-a|m or not-a|n).
Whether the inverse, converse or contrapositive holds can depend upon how you consider the qualifiers. I think the text intends the first simple version.
In that case consider when x = y = 0.