Let $$\begin{array}{rcl}a_1&=&25 \\ a_n&=&\underbrace{22\ldots 2}_{n \textrm{ times}}\underbrace{44\ldots 4}_{n-1 \textrm{ times}}5, \qquad n \geq 2. \end{array}$$
Prove that $a_n$ can be written as the sum of two perfect squares for any integer $n \geq 1$.
Proposed by Alessandro Ventullo, Milan, Italy
Solution:
Clearly, $a_1=9+16=3^2+4^2$. Moreover, we have $$\begin{array}{rcl} a_n&=&\underbrace{22\ldots 2}_{n \textrm{ times}}\underbrace{44\ldots 4}_{n-1 \textrm{ times}}5 \\ &=& \underbrace{22\ldots 2}_{n \textrm{ times}}\cdot 10^n + \underbrace{44\ldots 4}_{n-1 \textrm{ times}}\cdot10+5 \\ &=& \dfrac{2\cdot(10^n-1)}{9}\cdot10^n+\dfrac{4(10^{n-1}-1)}{9}\cdot10+5 \\ &=& \dfrac{2\cdot10^{2n}-2\cdot10^n+4\cdot10^n+5}{9} \\ &=&\dfrac{10^{2n}-2\cdot10^n+1}{9}+\dfrac{10^{2n}+4\cdot10^n+4}{9} \\ &=& \left(\dfrac{10^n-1}{3}\right)^2+\left( \dfrac{10^n+2}{3} \right)^2 \end{array}.$$ Observe that $\dfrac{10^n-1}{3}$ and $\dfrac{10^n+2}{3}$ are integers since $10^n-1$ and $10^n+2$ are divisible by $3$ for any $n \geq 1$.
No comments:
Post a Comment