Alice sums \(n\) consecutive numbers, not necessarily starting from \(1\), where \(n\) is a multiple of four. An example of such a sum is \(5+6+7+8\). Can this sum ever be odd?
Show that the difference between two consecutive square numbers is always odd.
One of the most powerful ideas in mathematics is that we can use letters — like \(a, b, n\) or \(x\) — to stand for numbers, shapes or other things. When we do this, we can reason about all such possible objects at once, without knowing exactly which number or shape we are really dealing with.
For example, the statement \[\text{``Let } a,b \text{ be numbers. Then } a+b=b+a."\] is true no matter what numbers \(a\) and \(b\) are. It tells us all of the following at the same time: \[3+5=5+3, \qquad (-10)+(-2)=(-2)+(-10), \qquad 7+0=0+7,\] and many more.
The rule \(a+b=b+a\) does not depend on the “three-ness” of \(3\) or the “five-ness” of \(5\) — it works for any numbers. Of course, we could not let \(a\) be a triangle and \(b\) be a tiger, because we do not know what it means to add a triangle to a tiger! Our rules only apply to objects for which the operations make sense.
This way of using symbols to express rules and patterns is what we call algebra. As long as we follow the rules that numbers follow, our reasoning will stay true. Today we will practise using these symbols to work with the algebra of numbers — it may take effort, but it is an important skill that will help you a lot in your mathematical journey.
Let \(n\) be a natural number and \(x=2n^2+n\). Prove that the sum of the square of the \(n+1\) consecutive integers starting at \(x\) is the sum of the square of the \(n\) consecutive integers starting at \(x+n+1\).
For example, when \(n=2\), we have \(10^2+11^2+12^2=13^2+14^2\)!
Find the contrapositive of the statement: “If in every school there is a class with at least \(20\) students, then there is a school with at least \(10\) students".
Show that if \(a\) and \(b\) are numbers, then \(a^2-b^2=(a-b)\times (a+b)\).
Show that if \(x,y,z\) are distinct nonzero numbers such that \(x+y+z = 0\), then we have \[\left(\frac{x-y}{z}+\frac{y-z}{x}+\frac{z-x}{y}\right)\left(\frac{z}{x-y}+\frac{x}{y-z}+\frac{y}{z-x}\right) = 9.\]
The Chinese remainder theorem is a fundamental result in number theory that allows one to decompose congruence problems to into simpler ones. The theorem says the following.
Suppose that \(m_1,m_2\) are coprime (i.e: they have no prime factors in common) natural numbers and \(a_1,a_2\) are integers. Then there is a unique integer \(x\) in the range \(0\leq x \leq m_1m_2-1\) such that \[x \equiv a_1 \pmod{m_1} \quad \text{ and } \quad x \equiv a_2 \pmod{m_2},\] where the notation \(x\equiv y \pmod{z}\) means that \(x-y=kz\) for some integer \(k\). Prove the Chinese remainder theorem using the pigeonhole principle.
We have ten positive integers \(x_1,\dots,x_{10}\) such that \(10\leq x_i\leq 99\) for \(1\leq i\leq 10\). Prove that there are two disjoint subsets of \(x_1,\dots,x_{10}\) with equal sums of their elements.
Let \(P(x)\) be a polynomial with integral coefficients. Suppose there exist four distinct integers \(a,b,c,d\) with \(P(a) = P(b) = P(c) = P(d) = 5\). Prove that there is no integer \(k\) with \(P(k) = 8\).