There are \(100\) people in a room, and each person has at least one friend in the room. Prove that amongst them there are two people with the same number of friends in the room (we don’t count being friends with oneself).
Can we obtain the polynomial \(h(x)=x\) by adding, subtracting, or multiplying the polynomials \(f(x)=x^2+x\) and \(g(x)=x^2+2\)?
The polynomial \(P(x)=x^3+3x^2-7x+1\) has three distinct roots: \(a,b,\) and \(c\). What is the value of \(a^2+b^2+c^2\)?
Find the remainder of dividing \(x^{100}-2x^{51}+1\) by \(x^2-1\). Try not to do a long calculation.
Katie and James take turns replacing the stars with any number of their choice in the following polynomial from left to right: \[x^4 +\star x^3+\star x^2+\star x+\star.\] Katie wins if once the game is over, the resulting polynomial has no integer roots (i.e: no roots which are whole numbers). Does James have a winning strategy?
Show that if a polynomial \(P(x)\) has an integer root (i.e: a root that is a whole number), then it can’t be that \(P(0)\) and \(P(1)\) are both odd.
Let \(x,x',y,y'\) be integers such that \(x+\sqrt{d}y=x'+\sqrt{d}y'\), where \(d\) is a number that is not a square. Show that \(x=x'\) and \(y=y'\).
Show that if \(u_1\) and \(u_2\) are solutions to Pell’s equation, then \(u_1u_2\) is also a solution to Pell’s equation. What can you conclude about the number of solutions, if there are any?
You have an \(8\times 8\) chessboard coloured in the usual way. You can pick any \(2\times 1\) or \(1\times 2\) piece and flip the white tiles to black tiles and vice-versa. Is it possible to finish with \(63\) white pieces and \(1\) black piece?
We start with the point \(S=(1,3)\) of the plane. We generate a sequence of points with coordinates \((x_n,y_n)\) with the following rule: \[x_0=1,y_0=3\qquad x_{n+1}=\frac{x_n+y_n}{2}\qquad y_{n+1}=\frac{2x_ny_n}{x_n+y_n}\] Is the point \((3,2)\) in the sequence?