### Proof

Let us consider natural numbers \(B\geq 2\) and \(D\), the representations of the naturals in base \(B\) and the transformation \(f_{D,B}: N_{D} \rightarrow N_{D}\) previously defined. One has:

(a) There is a natural \(D\) such that the function \(f_{D,B}\) has non-null fixed points if and only if \(B\) is not congruent to \(1\) modulo \(3\).

(b) If \(f_{D,B}\) has a non-null fixed point, then it has a representation \(a_{D-1}a_{D-2}...a_{1}a_{0}\) in base \(B\) verifying:

- if \(B\) is a multiple of \(3\), then \(a_{D-1} = 0\) and \(a_{0} = 0\), or \(a_{D-1} = \frac{B}{3}\) and \(a_{0} = \frac{2B}{3}\);

- if \(B\) is congruent to \(2\) modulo \(3\), then \(a_{D-1}=0\) and \(a_{0}=0\), or \(a_{D-1}=\frac{B-2}{3}\) and \(a_{0}=\frac{2B-1}{3}\).

In particular, in base \(10\), \(f_{D}\) has only one fixed point, \(0\), for every \(D\).

Let us fix a natural \(B\geq 2\) (a base) and an even natural \(D\). The fixed points of the transformation \(f_{D,B}\) are the solutions \(n\) of the equation \(f_{D,B}(n)=n\), or, equivalently, of the equations \(i_{D}(n)=n\) and \(i_{D}(n)=2n\). Let \(a_{D-1}a_{D-2}...a_{1}a_{0}\) be the representation of \(n\), in base \(B\), with \(D\) digits of \(\{0,1,...,B-1\}\). These equations are equivalent to the next two equalities: \[(B^{D-1} - 1)(a_{D-1}-a_{0}) + (B^{D-2} - B)(a_{D-2}-a_{1}) +...+\left (B^{\frac{D}{2}} - B^{\frac{D}{2}-1}\right)\left(a_{\frac{D}{2}}-a_{\frac{D}{2}-1}\right) =\\ B^{D-1}a_{D-1} + B^{D-2}a_{D-2} + ... + B^{\frac{D}{2}}a_{\frac{D}{2}} +...+ Ba_{1} + a_{0}\] or \[(B^{D-1} - 1)(a_{0}-a_{D-1}) + (B^{D-2} - B)(a_{1}-a_{D-2}) +...+\left (B^{\frac{D}{2}} - B^{\frac{D}{2}-1}\right)\left(a_{\frac{D}{2}-1}-a_{\frac{D}{2}}\right) =\\ B^{D-1}a_{D-1} + B^{D-2}a_{D-2} + ... + B^{\frac{D}{2}}a_{\frac{D}{2}} +...+ Ba_{1} + a_{0}.\]

The first equation is equivalent to \[a_{D-1} + Ba_{D-2} + ... + B^{D-2}a_{1} + B^{D-1}a_{0} = 0\] and from there we conclude that \(a_{D-1}=0=a_{D-2}=...=a_{1}=a_{0}\), thus obtaining the fixed point \(0\) of \(N_{D}\). The second equation is equivalent to \(i_{D,2}(x)=2x\) and it can be rewritten as \[B^{D-1} a_{0} + B^{D-2} a_{1} + ... + B a_{D-2} + a_{D-1} = \\B^{D-1}(2a_{D-1}) + B^{D-2}(2a_{D-2}) +...+ B^{\frac{D}{2}}\left(2a_{\frac{D}{2}}\right) +...+ B(2a_{1}) + 2a_{0}. \mbox{(#)}\]

We hence have an equality between a representation in base \(B\), on the left side, and a sum of powers of \(B\) between \(0\) and \(D-1\) with coefficients which are two times the coefficients of \(n\), on the right side. We then immediately deduce some necessary conditions for \(n\) to be a fixed point:

- \(B\) must divide \(2a_{0} - a_{D-1}\);
- if \(n\) is a non-null fixed point, then the reverse \(i_{D}(n)\) must be an even natural (in base \(10\));
- if \(a_{0} = 0\), then \(a_{D-1} = 0\);
- \(a_{D-1} < \frac{B}{2}\) (otherwise we would have a remainder from \(2a_{D-1}\) and, therefore, a nonzero summand with the power \(B^{D}\) in the right side of the equality);
- if \(a_{0} \neq 0\), then

- if \(2a_{0}\) has more than two
digits in base \(B\), then the digit \(a_{D-1}\) is the last digit of the
representation of \(2a_{0}\) in base \(B\);

- \(2a_{D-1} = a_{0}\) or \(2a_{D-1} +
1 = a_{0}\).

From the first and the last of these conditions, we also deduce that:

- Either \(a_{0}\) is even, equal to \(2a_{D-1}\), and \(B\) divides \(3a_{D-1}\).

In this case, since \(0\leq a_{D-1}\leq B-1\) and \(a_{D-1} < \frac{B}{2}\), we have only two possibilities for the equation \(3a_{D-1} = Bk\) with \(k\) a non-negative integer (since it implies \(0\leq k<\frac{3}{2}\)): either \(k=0\), and then \(a_{D-1} = 0\) and \(a_{0} = 0\); or \(k=1\) and \(a_{D-1} = \frac{B}{3}\), which is only possible if \(B\) is a multiple of \(3\), and \(a_{0} = \frac{2B}{3}\).

- Or \(a_{0}\) is odd, equal to \(2a_{D-1} + 1\), and \(B\) divides \(3a_{D-1} + 2\).

In this case, \(B\) can not be a multiple of \(3\), otherwise the prime \(3\) would divide \(3a_{D-1} + 2\), and would therefore be a divisor of \(2\). Moreover, since \(0\leq a_{D-1}\leq B-1\) and \(a_{D-1} < \frac{B}{2}\), only one nonnegative integer value \(k\) satisfies the equality \(3a_{D-1} + 2 = Bk\): \(k=1\) and \(a_{D-1}=\frac{B-2}{3}\), which is only possible if \(B\) is congruent to \(2\) modulo \(3\), and, therefore, \(a_{0} = \frac{2B-1}{3}\). Indeed, \(0\leq k\leq 2\) (because we should have \(\frac{3B}{2} + 2 > Bk\), that is, \(k < \frac{3}{2} + \frac{2}{B} \leq \frac{3}{2} + \frac{2}{2} = \frac{5}{2}\)); and, if \(k=0\), the equality \(3a_{D-1} + 2 = Bk\) is not possible; if \(k=2\), then \(a_{D-1} = 2\frac{B-1}{3}\), an equality which is valid only if \(B\) is congruent to \(1\) modulo \(3\) (whence \(B\geq 4\)), and \(a_{0} = \frac{4B-1}{3} = B + \frac{B-1}{3} \geq B + 1\), which is not allowed to a digit in base \(B\).

Let us see two examples. Consider \(D=2\) and \(B=2\). Since \(B\) is congruent to \(2\) modulo \(3\), the previous reasoning indicates how to build a non-null fixed point \(a_{1}a_{0}\): it is enough to consider \(a_{1} = \frac{B-2}{3} = 0\) and \(a_{0} =\frac{2B-1}{3} = 1\). And indeed \(01\) is a fixed point of \(f_{2,2}\). Let us now fix \(D=2\) and \(B=3\). The previous reasoning indicates that if the representation \(a_{1}a_{0}\) in base \(B\) is a fixed point, then \(a_{1} = \frac{B}{3} = 1\) and \(a_{0} = \frac{2B}{3} = 2\). But \(12\) (in base \(3\)) is not a fixed point because the reverse of \(n\) is \(21\), which is odd (in base \(10\)). When \(D=2\) and \(B=6\), a candidate to a fixed point is \(24\) (in base \(6\)), but, despite being even, it does not verify the equality (#). In general, if \(B\) is a multiple of \(3\), then \(f_{2,B}\) does not have non-null fixed points because such points would have to be represented by \(a_{1}a_{0}\), with \(a_{1} = \frac{B}{3}\) and \(a_{0} = \frac{2B}{3}\), and verify the equality (#), which in this case is \[B a_{0} + a_{1} = B\,(2a_{1}) + 2a_{0}\] that is, \[ B\frac{2B}{3} + \frac{B}{3} = B\frac{2B}{3} + \frac{4B}{3}\] \[\frac{1}{3}= \frac{4}{3}.\]

On the contrary, when \(B\) is congruent to \(2\) modulo \(3\), the transformation \(f_{2,B}\) has one (and only one) non-null fixed point: it is represented in base \(B\) by \(a_{1}a_{0}\), with \(a_{1} = \frac{B-2}{3}\) and \(a_{0} = \frac{2B-1}{3}\), which verifies equality (#) since \[\begin{array}{cl} & Ba_{0}+a_{1}=B\,\left(2a_{1}\right)+2a_{0}\\ \Leftrightarrow & B\frac{2B-1}{3}+\frac{B-2}{3}=B\left(2\frac{B-2}{3}\right)+2\frac{2B-1}{3}\\ \Leftrightarrow & 0=0 \end{array}\]

The condition that \(B\) is not congruent to \(1\) modulo \(3\), which is necessary for the existence of some even \(D\) such that \(f_{D,B}\) has a fixed point distinct of \(0\), is also a sufficient condition. Let us consider such a natural \(B\geq 2\); it only remains to guarantee the existence of such a fixed point for values of \(B\) which are multiples of \(3\). Let us search it for \(f_{4,B}\).

Let \(a_{3}a_{2}a_{1}a_{0}\) be an element of \(N_{4}\) in base \(B\). To verify whether it is a fixed non-null point, we must analyze one of the following cases.

Case 1: \(a_{3} = 0\) and \(a_{0} = 0\).

Equality (#) \[B^{3} a_{0} + B^{2} a_{1} + B a_{2} + a_{3} = \\ B^{3}\times 2a_{3} + B^{2}\times 2a_{2} + B\times 2a_{1} + 2a_{0}\] is reduced to \[B^{2}a_{1} + B a_{2} = B^{2}\times 2a_{2} + B\times 2a_{1}\] that is, \[B a_{1} + a_{2} = B\times 2a_{2} + 2a_{1} \] an equation which, in order that one obtains a non-null fixed point, imposes that \(a_{1} = \frac{2B}{3}\) and \(a_{2} = \frac{B}{3}\). However, as we have seen, these values do not satisfy the equality (#).

Case 2: \(a_{3} =\frac{B}{3}\) and \(a_{0} = \frac{2B}{3}\).

The equality (#) reads now as \[B^{3} \frac{2B}{3} + B^{2} a_{1} + B a_{2} + \frac{B}{3} =\\ B^{3}\frac{2B}{3} + B^{2}\times 2a_{2} + B\times 2a_{1} + \frac{4B}{3}\] that is, \[B^{2} a_{1} + B a_{2} = B^{2}\times 2a_{2} + B\times 2a_{1} + B\] which is reduced to \[B a_{1} + a_{2} = B\times 2a_{2} + 2a_{1} + 1.\]

This equation can now be analyzed as (#), thus obtaining

- \(B\) must divide \(2a_{1} + 1 - a_{2}\);
- \(a_{1}\neq 0\);
- \(a_{2} < \frac{B}{2}\);
- \(2a_{2} = a_{1}\) or \(2a_{2} + 1 = a_{1}\).

From the first and last of these conditions we conclude that:

- Either \(a_{1}\) is even, equal to \(2a_{2}\), and \(B\) divides \(3a_{2} + 1\).

But, since \(0\leq a_{D-2}\leq B-1\) and \(a_{2} < \frac{B}{2}\), the equality \(3a_{2} + 1 = Bk\) for a nonnegative integer \(k\) implies that \(k < \frac{3}{2} + \frac{1}{B} \leq \frac{3}{2} + \frac{1}{2} = 2\), from which it follows that \(k\) may be equal to \(0\) (a value which is not allowed because \(3a_{2} + 1\) is not zero) or \(k=1\) (a value which is not allowed because, by hypothesis, \(B\) is not congruent to \(1\) modulo \(3\)).

- Or \(a_{1}\) is odd, equal to \(2a_{2} + 1\), and \(B\) divides \(3a_{2} + 3\).

Since \(0\leq a_{D-2}\leq B-1\) and \(a_{D-2} < \frac{B}{2}\), we only have one admissible value for a nonnegative integer \(k\) satisfying the equality \(3a_{2} + 3 = Bk\): \(k=1\), and therefore \(a_{2} = \frac{B}{3}-1\) and \(a_{1} = \frac{2B}{3}\). That is, \(f_{4,B}\) has one (and only one) non-null fixed point, with representation in base \(B\) given by \[\left(\frac{B}{3}\right)\left(\frac{B}{3}-1\right)\left(\frac{2B}{3}-1\right)\left(\frac{B}{3}\right).\]

For example, if \(B=6\), then the number \(2134\) is the non-null fixed point of \(f_{4,6}\).