91Ó°ÊÓ

First, we need to prove that the equation is true when \(n = 1\). For n = 1, we have \(x^{2} - y^{2}\), which we can rewrite as \((x + y)(x - y)\). Since this expression is a product of \(x+y\) and another term, it is clearly divisible by \(x+y\). The base case is true.

Now we assume it is true for \(n = k\). That is, we assume that: \(x^{2k} - y^{2k} = (x+y)M\) for some integer M.

We need to prove that the statement is also true for \(n = k+1\), meaning we must show: \(x^{2(k+1)} - y^{2(k+1)} = (x+y)N\) for some integer N. Let's consider the expression for \(n = k+1\): \(x^{2(k+1)} - y^{2(k+1)} = x^{2k+2} - y^{2k+2}\) Now, let's rewrite this expression using a clever trick: \(x^{2k+2} - y^{2k+2} = x^{2}x^{2k} - y^{2}y^{2k}\) Factor the left side using the difference of squares formula: = \((x^{2} - y^{2})x^{2k} + y^{2}(x^{2k} - y^{2k})\) = \((x+y)(x-y)x^{2k} + y^{2}(x^{2k} - y^{2k})\) We know from our assumption that \(x^{2k} - y^{2k} = (x+y)M\). Additionally, the base case showed that \((x^2 - y^2) = (x+y)(x-y)\). So we can substitute these back into the equation: = \((x+y)(x-y)x^{2k} + y^{2}(x+y)M\) Now we can factor out the \((x+y)\) term: = \((x+y)[(x-y)x^{2k} + y^{2}M]\) Since the expression is now a product of \(x+y\) and another term, it is divisible by \(x+y\). We've proven that if the statement holds for \(n=k\), it also holds for \(n=k+1\). Thus, by mathematical induction, we have proven that \(x^{2n} - y^{2n}\) is divisible by \(x+y\) for all positive integers n.