NT-B2-M11-P015 Infinitely Many Automorphic Numbers

For each \(k\), solve \(x^2\equiv x\pmod{10^k}\) using the moduli \(2^k\) and \(5^k\).

The condition that the square ends with the same \(k\) digits is \(x^2\equiv x\pmod{10^k}\), or \(x(x-1)\equiv0\pmod{10^k}\). Since \(10^k=2^k5^k\), and \(x\) and \(x-1\) are coprime, the full powers \(2^k\) and \(5^k\) must be assigned to one of the two consecutive factors.

For example, the system \(x\equiv0\pmod{2^k}\), \(x\equiv1\pmod{5^k}\) has a unique solution modulo \(10^k\) by CRT. It is not simultaneously \(0\) and \(1\). For different large \(k\), these give new nonzero, non-one residues; otherwise one fixed integer would satisfy \(x^2\equiv x\pmod{10^k}\) for infinitely many \(k\), which is possible only for \(x=0\) or \(x=1\). Hence there are infinitely many automorphic numbers.

For class use, the solution can be made more concrete by constructing the first endings: \(76,376,9376,\ldots\).