Suppose that the last two digits $p$ and $q$ of $M$ and $N$ are not zero. It means that $p+q=10$.
Case 1: $p\ne q$
Let us first consider the case where $p\ne q$. I will use concrete values for $p,q$ but you can replace them with any other two. Suppose $p=6$, $q=4$.
The numbers $M,N$ are:
$$.........6 \\
.........4$$
Now, number 4 has to appear somewhere in the first number:
$$..4......6 \\
.........4$$
Under that number we must have digit 5 in the second number:
$$..4......6 \\
..5......4$$
Now you have 5 in the second number, it has to appear somewhere in the first:
$$..4.5....6 \\
..5......4$$
The matching number under it has to be 4:
$$..4.5....6 \\
..5.4....4$$
But now you have two 4s in the second number so you have to add one more to the first number and again add one 5 below it:
$$..4.5..4.6 \\
..5.4..5.4$$
But now you have the wrong count of 5s... and you are clearly in the infinite loop that you cannot exit.
Case 2: $p=q=5$
We have ten digit numbers both ending in 5:
$$.........5\\
.........5
$$
We have to fill 9 remaining positions in each number.
Missing digits come in pairs with their sum equal to 9 (digits cannot be equal, obviously). For example:
$$.3.......5\\
.6.......5
$$
But to keep the same digits in both numbers you have to put 6 into the first number and 3 into the last number:
$$.3..6....5\\
.6..3....5
$$
We have 7 positions left. Pick any two digits and you will have only five empty positions left:
$$.3..62.7.5\\
.6..37.2.5
$$
Eventually, you will end up with two incomplete numbers with all the same digits, just in a different order, and one empty position in each one. Whatever you choose to put there will break the "symmetry", your numbers won't have the same collection of digits.
Based on these two cases, we have a conclusion: the only way to construct the requested numbers is to start with $p=q=0$.
Best Answer
11 is congruent to 3 mod 4. While squares are only congruent 0 and 1 mod 4. Therefore $x^2-3y^2$ can only be $1-3,0-3,1-0$or $0-0$ So it can't be.
Proof squares are only conguent 1 and 0:
if k is even then $k=2m\rightarrow k^2=(2m)^2=4m^2$
if k is odd then $k=2m+1 \rightarrow k^2=(2m+1)^2=4m^2+4m+1=4(m^2+m)+1$