We prove that there exists a sequence $\sequence {b_n}$ with the properties: :$b_n < 10^{m n}$ :$b_n^2 \equiv \underbrace {\sqbrk {1 \paren {a_1 \cdots a_m} \cdots \paren {a_1 \cdots a_m}}}_{n \text { occurrences}} \pmod {2 \times 10^{m n}}$ by induction: === Basis for the Induction === For $n = 1$, we choose the number $b_1 = x \pmod {10^m}$ with $b_1 < 10^m$. Note that: {{begin-eqn}} {{eqn | l = b_1^2       | r = \paren {x - k \times 10^m}^2       | c = for some $k \in \Z$ }} {{eqn | r = k^2 \times 10^{2 m} - 2 x k \times 10^m + x^2       | c = Square of Sum }} {{eqn | o = \equiv       | r = x^2       | rr = \pmod {2 \times 10^m} }} {{eqn | o = \equiv       | r = \sqbrk {1 a_1 a_2 \cdots a_m}       | rr = \pmod {2 \times 10^m}       | c = by assumption }} {{end-eqn}} This is the basis for the induction. === Induction Hypothesis === This is our induction hypothesis: :There exists some $b_r$ such that: ::$b_r < 10^{m r}$ ::$b_r^2 \equiv \underbrace {\sqbrk {1 \paren {a_1 \cdots a_m} \cdots \paren {a_1 \cdots a_m}}}_{r \text { occurrences}} \pmod {2 \times 10^{m r}}$ Now we need to show true for $n = r + 1$: :There exists some $b_{r + 1}$ such that: ::$b_{r + 1} < 10^{m \paren {r + 1} }$ ::$b_{r + 1}^2 \equiv \underbrace {\sqbrk {1 \paren {a_1 \cdots a_m} \cdots \paren {a_1 \cdots a_m}}}_{r + 1 \text { occurrences}} \pmod {2 \times 10^{m \paren {r + 1}}}$ === Induction Step === This is our induction step: Let $b < 10^m$ and $b_{r + 1} = b \times 10^{m r} + b_r$. Note that: {{begin-eqn}} {{eqn | l = b_{r + 1}^2       | r = \paren {b \times 10^{m r} + b_r}^2 }} {{eqn | r = b^2 \times 10^{2 m r} + 2 b b_r \times 10^{m r} + b_r^2       | c = Square of Sum }} {{eqn | o = \equiv       | r = 2 b b_r \times 10^{m r} + \underbrace {\sqbrk {1 \paren {a_1 \cdots a_m} \cdots \paren {a_1 \cdots a_m} } }_{r \text { occurrences} } + 2 k \times 10^{m r}       | rr = \pmod {2 \times 10^{m \paren {r + 1} } }       | c = Induction Hypothesis; for some $k \in \Z$ }} {{end-eqn}} The rightmost $m r$ digits already satisfy the condition, so we consider the next $m + 1$ digits. We want: :$2 b b_r + 1 + 2 k \equiv \sqbrk {1 a_1 \cdots a_m} \pmod {2 \times 10^m}$ We first take Modulo $10^m$: :$2 b b_r + 1 + 2 k \equiv \sqbrk {a_1 \cdots a_m} \pmod {10^m}$ So we need to solve: :$b b_r + t \times 10^m = \dfrac {\sqbrk {a_1 \cdots a_m} - 1} 2 - k$ for integer solutions $b, t$. Since $a_m \ne 5$ and is odd: :$2, 5 \nmid b_r$ so $b_r$ and $10^m$ are coprime. By Bezout's Lemma, a solution for $b, t$ exists. We can also find a solution with $0 \le b < 10^m$. For this $b$, if; :$2 b b_r + 1 + 2 k \equiv \sqbrk {1 a_1 \cdots a_m} \pmod {2 \times 10^m}$ then take $b' = b$, $b_{r + 1} = b' \times 10^{m r} + b_r$ and we are done. It may happen that: :$2 b b_r + 1 + 2 k \equiv \sqbrk {0 a_1 \cdots a_m} \pmod {2 \times 10^m}$ In this case, take $b' = b \pm 5 \times 10^{m - 1}$, whichever is between $0$ and $10^m$. We have: {{begin-eqn}} {{eqn | l = 2b' b_r + 1 + 2 k       | r = 2 b_r \paren {b \pm 5 \times 10^{m - 1} } + 2 k + 1 }} {{eqn | r = 2 b_r b + 2 k + 1 \pm 10^m  }} {{eqn | r = \equiv \sqbrk {0 a_1 \cdots a_m} \pm 10^m       | rr = \pmod {2 \times 10^m} }} {{eqn | r = \equiv \sqbrk {1 a_1 \cdots a_m}       | rr = \pmod {2 \times 10^m} }} {{end-eqn}} hence $b_{r + 1} = b' \times 10^{m r} + b_r$ satisfy our conditions as well. By the Principle of Mathematical Induction, the sequence $\sequence {b_n}$ exists. {{qed}}