MathDB
Infinite base of a prime number!

Source: Iran 3rd round 2011-Number Theory exam-P5

September 19, 2012
modular arithmeticnumber theory proposednumber theory

Problem Statement

Let pp be a prime number. We know that each natural number can be written in the form ∑i=0taipi(t,ai∈N∪{0},0≤ai≤p−1)\sum_{i=0}^{t}a_ip^i (t,a_i \in \mathbb N\cup \{0\},0\le a_i\le p-1) Uniquely.
Now let TT be the set of all the sums of the form ∑i=0∞aipi(0≤ai≤p−1).\sum_{i=0}^{\infty}a_ip^i (0\le a_i \le p-1).
(This means to allow numbers with an infinite base pp representation). So numbers that for some N∈NN\in \mathbb N all the coefficients ai,i≥Na_i, i\ge N are zero are natural numbers. (In fact we can consider members of TT as sequences (a0,a1,a2,...)(a_0,a_1,a_2,...) for which ∀i∈N:0≤ai≤p−1\forall_{i\in \mathbb N}: 0\le a_i \le p-1.) Now we generalize addition and multiplication of natural numbers to this set so that it becomes a ring (it's not necessary to prove this fact). For example:
1+(∑i=0∞(p−1)pi)=1+(p−1)+(p−1)p+(p−1)p2+...1+(\sum_{i=0}^{\infty} (p-1)p^i)=1+(p-1)+(p-1)p+(p-1)p^2+... =p+(p−1)p+(p−1)p2+...=p2+(p−1)p2+(p−1)p3+...=p+(p-1)p+(p-1)p^2+...=p^2+(p-1)p^2+(p-1)p^3+... =p3+(p−1)p3+...=...=p^3+(p-1)p^3+...=...
So in this sum, coefficients of all the numbers pk,k∈Np^k, k\in \mathbb N are zero, so this sum is zero and thus we can conclude that ∑i=0∞(p−1)pi\sum_{i=0}^{\infty}(p-1)p^i is playing the role of −1-1 (the additive inverse of 11) in this ring. As an example of multiplication consider (1+p)(1+p+p2+p3+...)=1+2p+2p2+⋯(1+p)(1+p+p^2+p^3+...)=1+2p+2p^2+\cdots
Suppose pp is 11 modulo 44. Prove that there exists x∈Tx\in T such that x2+1=0x^2+1=0.
Proposed by Masoud Shafaei
Back to ProblemsView on AoPS