since we have $$n\equiv 0,1,2,3,4,5,6\mod 7$$ we get $$n^2\equiv 0,1,2,4\mod 7$$
therefore $$n^2+4\equiv 1,4,5,6\mod 7$$
Since you are concerned only with rectangular patterns, you have four digit numbers in a certain base $l$, and you want to check divisibility by $11$, where $11 = l+1$.
When you have a four digit number in base $l$, write it as $al^3 + bl^2 + cl + d$, where $0 \leq a,b,c,d < l$. (This represents the $l$-base number $\overline{abcd}$). Now, we have a cute fact : $l^3 + 1$ is a multiple of $l+1$, since $l^3+1 = (l^2 - l + 1)(l+1)$. Furthermore, $l^2 - 1 = (l-1)(l+1)$. Therefore, we make the following rewrite :
$$
al^3 + bl^2 + cl + d = a(l^3 + 1) + b(l^2 - 1) + c(l+1) - (a - b + c - d) \\ = (l+1)(...) + ((b+d)-(a+c))
$$
where $l+1 = 11$ in base $l$.
Therefore, the remainder when $\overline{abcd}$ is divided by $11$ is $(b+d) - (a+c)$.
When you consider four numbers ($1 \to 9$, since I found problems in the letters for hex) in a rectangle and form a four digit number out of them now, can you see why this number $(b+d) - (a+c)$ is in fact zero, therefore giving your desired result?
Now that we have pointed out this rectangular pattern, I noted above that some counterexamples did exist for the MS - layout of hexadecimal numbers. The issue there was a fairly simple one : the "matrix" of entries did not satisfy the property that $a+d = b+c$ for $a,b,c,d$ going CW/CCW around any $2\times 2$ subrectangle of entries of the matrix.
If $l-1$ is not a prime, then we can actually arrange a "matrix" of $l$ entries which is not strictly column or row, but satisfies this "rectangle property" as we can call it. For this, write $l-1 = ab$ where $a,b \neq 1$, and arrange an $a \times b$ matrix of entries, which we fill in the following fashion : start with $1$ in the bottom corner, proceed towards right filling $2,3,...$ until you hit the end, then return to the left of the row above, and fill the next number, now repeat till you fill the whole matrix.
For $10$, this procedure yields the conventional keyboard pattern. For $15 = 5 \times 3$, it would yield $$
\begin{pmatrix}
B&C &D & E& F\\
6&7&8&9&A\\
1&2&3&4&5 \\
\end{pmatrix}
$$
which indeed will satisfy the property that any rectangle has $11$ in that base as a divisor. For example, $A8DF$, $1496$ and $CE97$ are all multiples of $11$.
Note that more is true : in fact, every parallellogram , read CW or CCW, leads to a multiple of eleven.
Best Answer
Problems like this appear frequently here. There are a couple of standard approaches. One is to use Fermat's little theorem, which says that if $p$ is a prime number, then $n^p-n$ is divisible by $p$ for all $n$.
Since $42=2\times 3\times 7$, what we need to do is to check that 2, 3, and 7 divide $n^7-n$, no matter what $n$ is.
That 7 does is direct from Fermat's little theorem.
The theorem also ensures that 3 divides $n^3-n$. Now: $$n^7-n=n(n^6-1)=n((n^2)^3-1)=n(n^2-1)(n^4+n^2+1)=(n^3-n)(n^4+n^2+1),$$ so 3 indeed divides $n^7-n$.
Finally, $n$ and $n^7$ always have the same parity, so $n^7-n$ is even.
We can actually argue without Fermat's little theorem in this case. An approach that only requires patience is as follows: The idea is to factor the polynomial $x^7-x$ and then analyze the result when $x=n$ is an integer. (This is a trick that Bill Dubuque suggests sometimes in his solutions.)
We have: $x^7-x=x(x^6-1)=x(x^3+1)(x^3-1)=x(x+1)(x^2-x+1)(x-1)(x^2+x+1)$. When $x=n$, we have $$ n^7-n=(n-1)n(n+1)(n^2-n+1)(n^2+n+1). $$ Now we analyze this prime by prime, as before. Note that one of $n$ and $n-1$ is always even, so the product is even. Also, of 3 consecutive numbers, such as $n-1,n,n+1$, one is always divisible by 3, so it only remains to verify divisibility by 7.
We may assume that $n=7k+b$ where $b=\pm2$ or $\pm3$, since otherwise $(n-1)n(n+1)$ is a multiple of 7. In that case, $n^2\equiv 4$ or $2\pmod 7$, and one of $n^2+n$, $n^2-n$ is $\equiv 6\pmod 7$, so $(n^2-n+1)(n^2+n+1)$ is a multiple of 7.
The disadvantage of this approach over the previous one, of course, is the need to analyze different cases. Fermat's little theorem allows us to analyze all cases simultaneously, which typically (as here) results in a much faster approach.
If you are comfortable with the method of induction, this gives us a way of verifying divisibility by 7 which is not without some elegance (divisibility by 2 and 3 is probably best approached as before). Note that $(-n)^7-(-n)=-(n^7-n)$, so we may as well assume that $n\ge 0$. If $n=0$ it is obvious that 7 divides $n^7-n$.
Suppose then that $7|n^7-n$, and argue that $7|(n+1)^7-(n+1)$. For this, actually expand $(n+1)^7$ using the binomial theorem: $$ (n+1)^7=n^7+7n^6+{7\choose 2}n^5+{7\choose 3}n^4+\dots+1.$$ The point is that $${7\choose k}=\frac{7!}{k!(7-k)!}$$ is obviously divisible by 7 as long as $k\ne0,7$, so (modulo 7) we have that $(n+1)^7-(n+1)\equiv (n^7+1)-(n+1)=(n^7-n)$. Now we invoke the induction hypothesis, that precisely says that the latter is divisible by 7, and we are done.
Of course, exactly the same inductive argument gives us a proof of Fermat's little theorem: $p|n^p-n$ for any $p$ prime and any integer $n$.