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 if Katie starts first?
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 two adjacent squares (i.e: any \(2\times 1\) or \(1\times 2\) section of the board) 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 \((1,3)\) of the plane. We generate a sequence of points with the following rule: the \(x\)-coordinate of the new point is the arithmetic mean of the \(x\) and \(y\) coordinates of the previous point, and the \(y\)-coordinate of the new point is the harmonic mean of the \(x\) and \(y\) coordinates of the previous points. The harmonic mean of two numbers \(x\) and \(y\) is \(\frac{2}{\frac{1}{x} + \frac{1}{y}}\). Is the point \((3,2)\) in the sequence?
Four black dots are drawn on a whiteboard. On the dots we write the numbers \(10\), \(20\), \(30\), and \(40\) (one number on each dot). We then repeat the following move any number of times: choose one dot, decrease its number by \(3\), and increase the number on each of the other three dots by \(1\). After some number of moves, is it possible for all four dots to show the number \(25\) simultaneously?
Often in maths we want to prove statements of the form “If A, then B.” For example: “If a number is divisible by \(4\), then it’s even.”
Usually, we prove such statements using something called a direct proof. In the example above, a direct proof would start by imagining we have some number — we don’t know which one — but we know it has the property “divisible by \(4\)”, and then using this information to work out that the number must be even.
However, this kind of direct reasoning can sometimes be tricky. Luckily, there’s another way! The idea is that a statement of the form “If A, then B” means exactly the same thing as “If not \(B\), then not \(A\).”
This second way of writing it is called the contrapositive, and we call “not \(B\)” the negation of \(B\). Here’s an everyday example: “If it rains, then I take my umbrella." is exactly the same as saying “If I don’t take my umbrella, then it’s not raining."
When we use this method in maths, we often say we’re proving by contrapositive: instead of proving “If \(A\) then \(B\)”, we prove “If not \(B\) then not \(A\).” Using this idea, to prove the example above, it would be the same as to prove the statement: ”If a number is not even, then it’s not divisible by \(4\)".
We sometimes write “If \(A\) then \(B\)” as \(A \implies B\), which is pronounced “\(A\) implies \(B\)”. Its contrapositive is: \(\text{not }B \implies \text{not }A.\) This way of thinking often makes a proof much simpler. Let’s see some examples to learn how to use this method.