 # Understand the problem

The sequence of positive integers $a_1, a_2, a_3, ...$ satisfies $a_{n+1} = a^2_{n} + 2018$ for $n \ge 1$.
Prove that there exists at most one $n$ for which $a_n$ is the cube of an integer.

##### Source of the problem

Ireland MO 2018, Problem 9

Number Theory
8/10
##### Suggested Book
Excursion in Mathematics by Bhaskaryacharya Prathisthan

Do you really need a hint? Try it first!

, wIt is so important to know and use the modulo technqiue at the right time.  We will use the modulo technique, i.e. we will see the problem through the lens of modulo some number. What is that number? If you visit this website, you will understand that to handle cubes modulo something is 9. So, we will deal the whole equation modulo 9.

Definition: kth power residue of a number n is the complete residue system modulo n. For eg: Quadratic Residue (2nd power) of 4 is {0,1}.

• Cubic(3rd) Power Residue of 9 is {0,1,-1}.
• 6th Power Residue of 9 is {0,1}
• Quadratic(2nd Power) Residue of 9 is {0,1,4,7}
We will use these ideas here.
Let $a_k$ be the smallest integer which is a cube; let $a_k=a^3$. Note that, $a_{k+1}=a^6+2018$.  Now, the modulo picture comes in. Starting from this cube. We will observe the sequence modulo 9. Case 1: $a_k = 0$ mod 9 Then, the sequence modulo 9 will be $0 \mapsto 2 \mapsto 6 \mapsto 2 \mapsto \dots$ Hence, there are no further cubes possible as the cubic residues of 9  are {0,1,-1}.
Case 2: $a_k = 1,-1$ mod 9 Then, the sequence modulo 9 will be $\pm 1 \mapsto 3 \mapsto 2 \mapsto 6 \mapsto 2 \mapsto \dots$ Hence, there are no further cubes possible as the cubic residues of 9  are {0,1,-1}. QED

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