## Number Theory Homework (2 Weeks)

Homework problems for 5 Nov 2011:

1. Let a1, a2, … be a series recursively defined as follows: a1 = 20, a2 = 11, and for n ≥ 1, an+2 is the remainder when an+1 + an is divided by 100. What is the remainder when $a_1^2 + a_2^2 + \dots + a_{2011}^2$ is divided by 8?
2. (IMO 1962 Q1) Find the smallest positive integer with the following properties:
• in decimal representation, its last digit is 6;
• if we move this digit to the front of the number, we get 4 times the original number.
3. Find two 3-digit numbers (abc) and (def) such that upon placing them side-by-side, we get a 6-digit number (abcdef) satisfying : (abcdef) = ( (abc) + (def) )2.
4. (a) A positive integer is written on the board. At each turn, we erase its unit digit and add 4 times that digit to the remaining number. Starting from 52011, can we get 20115 at some point? (b) Suppose instead, at each turn, we erase its unit digit and add 5 times that digit to the remaining number. Starting from 72011, can we get 20117 at some point?
5. Let a1, a2, … be a series recursively defined: a0 = 1, a1 = 3, and for n ≥ 0, an+2 = 6an+1an. Prove that for each n, either $\frac{a_n+1} 2$ or $\frac{a_n-1}2$ is a perfect square.

Homework problems for 12 Nov 2011:

1. Prove that if a, n are integers and n > 0, then whenever $a \equiv 1 \pmod n$, we have $a^n \equiv 1 \pmod {n^2}$.
2. Classify the solutions of $\frac 1 x + \frac 1 y = \frac 1 z$, where x, y and z are positive integers.
3. Prove that if n =3k, then n | (2n + 1). In particular, there are infinitely many n such that n | (2n + 1). Can you find an n with at least 2 distinct prime factors such that n | (2n + 1)?
4. (IMO 1971 shortlisted) Find all integer solutions to $x^2 + y^2 = (x-y)^3$.
5. Suppose n is a positive integer such that n | (2n + 1). Prove that n = 1 or n is a multiple of 3.

[ Coming attractions : quadratic residues… stay tuned! 🙂 ]

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### 2 Responses to Number Theory Homework (2 Weeks)

1. Zhang Hui says:

Homework problems for 12 Nov 2011: Q5
something wrong with this question, if n=6, (multiple of 3), but n cannot divide 2^{n}+1.
n should be a power of 3, i.e, n=3^{k}, k\in \mathbb{N}_{0}

2. limsup says:

The implication is only one way: the converse doesn’t hold.