Find the minimum in a proper subset with those condtions

Short proof

We have $A = A + 19\mathbb Z$ and for any $b \in A$ , $A = A + 2b\mathbb Z$ . Thus $A = A + 19\mathbb Z + 2b\mathbb Z = A + \text{gcd}(19,2b)\mathbb Z,$ but since $A\not = \mathbb Z$ , we must have $\text{gcd}(19,2b) > 1$ so that $19 | b$ and $A \subseteq 19\mathbb Z$ . However, if $A \not= \emptyset$ then also $A \supseteq b + 19\mathbb Z = 19\mathbb Z$ , so that in fact $A = 19\mathbb Z$ and the desired minimum is $\boxed{19}$ .

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Longer version for those who don't like concise set notation.

(1) If $a \in A$ , then $a + 19n \in A$ for all integers $n$ . Proof:

• For $n = 0$ it is trivial.

• For $n > 0$ , by induction. If $a' = a + 19(n-1) \in A$ , then according to the rule, $a' + 19 = a + 19n \in A$ .

• For $n < 0$ , by induction. If $a' = a + 19(n+1) \in A$ , then $-a' \in A$ , so $-a' + 19 \in A$ , so $-(-a' + 19) = a' - 19 = a + 19n \in A$ .

(2) If $a,b \in A$ , then $a + 2bn \in A$ for all integers $n$ . Proof:

• Assuming $a \not = b$ , use a similar reasoning by induction as above.

• The case $a = b$ is easily avoided by observing that $a + 19 \in A$ , so $a + 19 + 2bn \in A$ for all $n$ , so $a + 19 + 2bn - 19 = a + 2bn \in A$ for all $n$ .

(3) Suppose $b \in A$ is not a multiple of 19, and suppose $a \in A$ . Let $c \in \mathbb Z$ . Then $c \in A$ . Proof:

• $\text{gcd}(2b,19) = 1$ .

• From the (extended) Eucleidan algorithm, there exist integers $n_1, n_2$ such that $c - a = 2bn_1 + 19n_2$ .

• Since $a \in A$ , we know from (2) that $a + 2bn_1 \in A$ , and from (1) that $a + 2bn_1 + 19n_2 = c \in A$ .

(4) We are told that $A \not= \mathbb Z$ , so that there must be an integer $c \not \in A$ . Using the converse of (3), we conclude that there does not exist $b \in A$ that is not a multiple of 19. It follows that $A$ only contains multiples of 19.

(5) Since $A$ is not empty, it contains at least one such multiple of 19, say $a = 19m$ . But from (1) we know that then all integers $a + 19n = 19(n+m)$ are elements of $A$ . In other words, $A$ is precisely the set of all multiples of 19 ( $A = 19\mathbb Z$ ).

It is now easy to conclude that the answer to the problem is $\boxed{19}$ .

Denote $m=\min \limits_{a \in A}(|a|)$ we first proof this claim:

Claim: $A$ is a subgroup of $(\mathbb Z, +)$

Let $a,b \in A$ we have : $(\forall m\ge1) : a+2mb \in A$ ( Just use induction, easy!) so $a+20b \in A$ . By easy induction we also have: $(\forall x\in A)(\forall n\in \mathbb Z) : x+19n \in A$ for $n=-b$ and $x=a+20b$ we find: $a+b \in A$ $\square$

$A$ is a subgroup of $(\mathbb Z, +)$ so $A= m\mathbb Z$ witch means that if $x \in A$ : $m | x$ but we have also $m | (x+19)$ since $x+19\in A$ so $m | 19$ . $m=19$ , since we cannot have $m=1$ because then we would have $A=\mathbb Z$ .

Assume that there is an element $b$ in $A$ that is not divisible by $19$ . Then $2b$ is also not divisible by $19$ , and by Fermat’s little theorem $(2b)^{18} \equiv 1 \mod 19$ . Then through multiple uses of the rule $a + 2b$ we can start with an element $a$ (not equal to $b$ ) and arrive at an element $c = a + (2b)^{18} \equiv a + 1 \mod 19$ , and through multiple uses of the rule $-a$ and $a + 19$ we can obtain element $d = a + 1$ .

In other words, if $b$ is not divisible by $19$ , we can carry out the above steps on element $a$ to obtain element $a + 1$ , and again on element $a + 1$ to obtain element $a + 2$ , and so on. Using this and the rule of $-a$ , all the integers $\mathbb Z$ can be obtained in $A$ . But this contradicts the fact that $A$ is a proper subset of $\mathbb Z$ .

Therefore, all the elements in $A$ must be divisible by $19$ , and the smallest non-zero multiple of $19$ is $\boxed{19}$ .