I tried to show that $$ {\left(\sqrt{2\,} + 1\right)^{2n+1} + \left(\sqrt{2\,} - 1\right)^{2n+1} \over 2\,\sqrt{2\,}}\,,\qquad n\geq2 $$ is written as the sum of two perfect squares. We used Newton's binomial formula and we did. Is there another way ?. Thanks you !.
Asked
Active
Viewed 86 times
1 Answers
8
Let us look at a quadratic whose roots are $$(\sqrt2 + 1)^2 \text{ and }(\sqrt2 - 1)^2$$ i.e., $$3+2\sqrt2 \text{ and }3 - 2\sqrt2$$ The quadratic is $x^2-6x+1$. Now let us look at the recurrence $$a_{n+1} = 6a_n - a_{n-1}$$ where $a_0 = 1$ and $a_1 = 5$. This gives the sequence you are after. Now note that $$a_0 = 0^2 + 1^2$$ $$a_1 = 1^2 + 2^2$$ $$a_2 = 2^2 + 5^2$$ $$a_3 = 5^2 + 12^2$$ $$a_4 = 12^2 + 29^2$$ Hence, it looks like if $a_{n-1} = x^2 + y^2$, then $a_n = y^2 + (2y+x)^2$. Prove this is the case using the recurrence and induction. The identity $$(2(2y+x) + y)^2 + (2y+x)^2 = 6(y^2 + (2y+x)^2) - (x^2+y^2)$$ will be helpful in the process.