Show that the difference between two consecutive square numbers is always odd.
A basic feature of mathematics is that we can argue in great generality. We do so by working with objects (numbers, shapes or other things) that are “generic" in some sense. We usually use letters and symbols such as \(a,b,n\) or \(x\) to refer to such objects.
The statement \[\text{``Let }a,b\text{ be numbers. Then we have }a+b=b+a\text{."}\] is completely general. It implies all of the following \[3+5=5+3, \quad (-10)+(-2) = (-2)+(-10), \quad 7+0 =0+7\] and much much more.
The statement \(a+b=b+a\) is of course true when \(a=3\) and \(b=5\), but it clearly does not depend on the “three-ness" of \(a\) or “five-ness" of \(b\). This is what we mean by generic. On the other hand, we couldn’t 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!
By using names to refer to a generic object, our statement and deductions become true for all objects satisfying the same hypothesis as our generic objects. The language of algebra provides a very efficient way for expressing relations and operations we perform on objects. As long as we follow the formal rules provided by algebra, our conclusion is always correct. Of course, the formal rules do not appear out of thin air - they capture some kind of intuition we have about the objects we are dealing with. In the case of the rule \(a+b=b+a\), this says the sum of two numbers should not depend on the order we add them as shown in the examples above.
The purpose of this sheet today is to work with the algebra of numbers. It is hard work, but it is an essential skill in mathematics and will pay off very handsomely in the long run!
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 \(10^2+11^2+12^2=13^2+14^2\)!