Let s and t be two different positive integers which are relatively prime. Does this property alone (of being relatively prime) NECESSARILY IMPLIES that one of s and t MUST be even and the other odd?
If yes, how would you prove so?
[Math] Number theory: two relatively prime numbers, necessarily one must be odd and the other even
number theory
Related Solutions
Yes. The generalization is provided by modular arithmetic. The properties you are observing all come from the fact that taking the remainder modulo $n$ respects addition and multiplication, and this generalizes to any $n$. More generally in abstract algebra we study rings and their ideals for the same reasons.
The notion of evenness and oddness of functions is closely related, but it is somewhat hard to explain exactly why. The key point is that there is a certain group, the cyclic group $C_2$ of order $2$, which is behind both concepts. For now, note that the product of two even functions is even, the product of an even and odd function is odd, and the product of two odd functions is even, so even and odd functions under multiplication behave exactly the same way as even and odd numbers under addition.
There are also huge generalizations depending on exactly what you're looking at, so it's hard to give a complete list here. You mentioned chessboards; there is a more general construction here, but it is somewhat hard to explain and there are no good elementary references that I know of. Once you learn some modular arithmetic, here is the modular arithmetic explanation of the chessboard idea: you can assign integer coordinates $(x, y)$ to each square (for example the coordinate of the lower left corner), and then you partition them into black or white squares depending on whether $x + y$ is even or odd; that is, depending on the value of $x + y \bmod 2$. Then given two points $(a, b)$ and $(c, d)$ you can consider the difference $c + d - a - b \bmod 2$, and constraints on this difference translate to constraints on the movement of certain pieces. This idea can be used, for example, to prove that certain chessboards (with pieces cut out of them) cannot be tiled with $1 \times 2$ or $2 \times 1$ tiles because these tiles must cover both a white square and a black square. Of course there are generalizations with $2$ replaced by a larger modulus and larger tiles.
As for matrices and vectors, let's just say that there are a lot of things this could mean, and none of them are straightforward generalizations of the above concept.
$1.$ By a theorem of Euler, any even perfect number is of the shape $2^{p-1}(2^p-1)$ where $p$ is prime. We don't need the primality part. Multiply by $8$, add $1$. We get $2^{2p+2}-2(2^{p+1})+1$, which is the square of $2^{p+1}-1$.
$2.$ We show that an odd perfect number cannot have only $2$ distinct prime factors (we don't deal with only $1$ prime factor, it's easier).
Let $N=p^aq^b$ where $p$ and $q$ are odd primes and $p\lt q$. Then the sum of the divisors of $N$ is $$(1+p+\cdots+p^a)(1+q+\cdots+q^b).$$ Using the ordinary formula for the sum of a finite geometric series, this can be rewritten as $$\frac{p^{a+1}-1}{p-1}\cdot \frac{q^{b+1}-1}{q-1}.$$ Divide by $N$. The result is $$\frac{p-\frac{1}{p^a}} {p-1}\cdot \frac{q-\frac{1}{q^b}} {q-1} .$$ This is less than $$\frac{p}{p-1}\cdot \frac{q}{q-1}.\tag{$1$}$$ We show that the product $(1)$ must be less than $2$. In particular, that shows the product cannot be $2$, so $N$ is in fact deficient.
It is easier to show that the reciprocal of Expression $(1)$ is greater than $\frac{1}{2}$. This reciprocal is $$\left(1-\frac{1}{p}\right)\cdot \left(1-\frac{1}{q}\right).$$
But $p\ge 3$ and $q\ge 5$. So $1-\frac{1}{3}\ge \frac{2}{3}$ and $1-\frac{1}{q}\ge \frac{4}{5}$. So their product is $\ge \frac{8}{15}$, which is greater than $\frac{1}{2}$.
I have not thought about your Question $3$.
Best Answer
No, for example $3$ and $5$ are relatively prime, with both being odd integers. For that matter, all primes, including $2$, are relatively prime to each other.