$3^2+2=11$, $33^2+22=1111$, $333^2+222=111111$, and so on.

Question

$3^2+2=11$

$33^2+22=1111$

$333^2+222=111111$

$3333^2+2222=11111111$

$\vdots$

The pattern here is obvious, but I could not have a proof.

Prove that $\underset{n\text{ }{3}\text{'s}}{\underbrace{333\dots3}}^2+\underset{n\text{ }{2}\text{'s}}{\underbrace{222\dots2}}=\underset{2n\text{ }{1}\text{'s}}{\underbrace{111\dots1}}$ for any natural number $n$.

Dear, I am not asking you to prove, I just want a hint, how to start proving it. Thanks.

Answer 1

Because $$\left(\frac{10^n-1}{3}\right)^2+2\cdot\frac{10^n-1}{9}=\frac{10^{2n}-2\cdot10^n+1+2\cdot10^n-2}{9}=\frac{10^{2n}-1}{9}.$$

Answer 2

Hint: $$\underbrace{aaa\cdots a}_{n\text{ a's}}=\frac{a(10^n-1)}{9}$$ Where $a\in\{0,1,2,3,4,5,6,7,8,9\}$ and $n\in\mathbb{N_0}$.

Answer 3

There's also a simple argument using only divisors.

small example: $$ 111111 - 222 = 111 \times 1001 - 111 \times 2 = 111 \times 999 = 333^2 $$

generalized: $$ \underbrace{11\cdots1}_{2n} - \underbrace{22\cdots 2}_{n} = \underbrace{11\cdot 1}_{n} \times \underbrace{100\cdots01}_{n+1} - 2 \times \underbrace{11\cdots1}_{n} = \underbrace{99\cdots 9}_{n} \times \underbrace{11 \cdots 1}_{n} = (\underbrace{33\cdots3}_{n})^2 $$

Answer 4

$${\underbrace{33...3}_{n}}^2+\underbrace{22...2}_{n}=\underbrace{33..3}_{n}\cdot \underbrace{11...1}_{n}\cdot 3+\underbrace{11...1}_{n}\cdot 2=\\ \underbrace{11...1}_{n}(\underbrace{99...9}_{n}+1+1)=\underbrace{11...1}_{n}\cdot 1\underbrace{00...0}_{n}+\underbrace{11...1}_{n}=\underbrace{11...1}_{n}\underbrace{00...0}_{n}+\underbrace{11...1}_{n}=\underbrace{11...1}_{2n}.$$