The proof proceeds by induction on $n$. For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition: :There are exactly $4$ $n$-digit automorphic numbers of the forms above. === Basis for the Induction === For $n = 1$: $0, 1, 5, 6$ are the only $1$-digit automorphic numbers, and we have: :$5^{2^0} = 5$ :$6^{5^0} = 6$ Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. We assume that for some $k \ge 1$, there are exactly $4$ $k$-digit automorphic numbers of the forms above. === Induction Step === This is the induction step: We aim to construct $x$, a $\paren {k + 1}$-digit automorphic number. By Left-Truncated Automorphic Number is Automorphic, after removing the leftmost digit, what remains is a $k$-digit automorphic number. Write $x = 10^k a + b$, where $a$ is the leftmost digit of $x$. We have: {{begin-eqn}} {{eqn | l = x^2       | r = \paren {10^k a + b}^2 }} {{eqn | r = 10^{2 k} a^2 + 2 \times 10^k a b + b^2       | c = Square of Sum }} {{eqn | r = 10^{2 k} a^2 + 2 \times 10^k a b + \paren {b + 10^k N}       | c = for some $N \in \N$; {{Defof|Automorphic Number}} }} {{eqn | o = \equiv       | r = 10^k \paren {2 a b + N} + b       | rr= \pmod {10^{k + 1} } }} {{eqn | o = \equiv       | r = x       | rr= \pmod {10^{k + 1} }       | c = {{Defof|Automorphic Number}} }} {{eqn | o = \equiv       | r = 10^k a + b       | rr= \pmod {10^{k + 1} } }} {{eqn | ll = \leadsto       | l = 2 a b + N       | o = \equiv       | r = a       | rr= \pmod {10} }} {{eqn | ll = \leadsto       | l = N       | o = \equiv       | r = a \paren {1 - 2 b}       | rr= \pmod {10} }} {{end-eqn}} By Left-Truncated Automorphic Number is Automorphic, $b$ must end in $0, 1, 5, 6$. We see that: :$a \equiv N \pmod {10}$ for $b$ ending in $0$ or $5$; :$a \equiv -N \pmod {10}$ for $b$ ending in $1$ or $6$; So the choice of $x$ is completely determined by $b$. This shows that there are exactly $4$ $\paren {k + 1}$-digit automorphic numbers. {{qed|lemma}} Now we show that the $4$ numbers are indeed of the forms given above. For any $n \in \Z_{>0}$, we have: {{begin-eqn}} {{eqn | l = 0^2 - 0       | o = \equiv       | r = 0       | rr= \pmod {10^n} }} {{eqn | l = 1^2 - 1       | o = \equiv       | r = 0       | rr= \pmod {10^n} }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \paren {5^{2^{n - 1} } }^2 - 5^{2^{n - 1} }       | r = 5^{2^{n - 1} } \paren {5^{2^n} - 1} }} {{eqn | r = 5^{2^{n - 1} } \paren {5 - 1} \paren {5 + 1} \paren {5^2 + 1} \dots \paren {5^{2^{n - 2} } + 1}       | c = Difference of Two Squares }} {{end-eqn}} Since $2^{n - 1} \ge n$ and the each factor on the right is divisible by $2$, the above product is divisible by $5^n 2^n = 10^n$. {{begin-eqn}} {{eqn | l = \paren {6^{5^{n - 1} } }^2 - 6^{5^{n - 1} }       | r = 6^{5^{n - 1} } \paren {6^{5^{n - 1} } - 1} }} {{eqn | r = 6^{5^{n - 1} } \paren {6^{5^{n - 2} } - 1} \paren {6^{4 \times 5^{n - 2} } + 6^{3 \times 5^{n - 2} } + 6^{2 \times 5^{n - 2} } + 6^{5^{n - 2} } + 1}       | c = Difference of Two Odd Powers }} {{eqn | r = 6^{5^{n - 1} } \paren {6^{5^{n - 2} } - 1} \paren {5 N}       | c = for some $N \in \N$: Congruence of Powers }} {{eqn | o = : }} {{eqn | r = 2^{5^{n - 1} } 3^{5^{n - 1} } \paren {6^{5^0} - 1} \paren {5^{n - 1} C}       | c = for some $C \in \N$ }} {{eqn | r = 2^{5^{n - 1} } 3^{5^{n - 1} } \paren {5^n C} }} {{end-eqn}} Since $5^{n - 1} \ge n$, the above product is divisible by $2^n 5^n = 10^n$. Hence for each number above: :$x^2 \equiv x \pmod {10^n}$ showing they are indeed automorphic. So $\map P k \implies \map P {k + 1}$ and thus it follows by the Principle of Mathematical Induction that these are all the automorphic numbers. {{qed}} {{expand|Add a proof via p-adic numbers}} Category:Automorphic Numbers jhzrvo7ggjkanrgbcquux60qq8dx0u8