1. Zadatak

Prvi način:

$$a-b=\underbrace{11\ldots11}_{2016}-\underbrace{222\ldots22}_{1008}=\frac{1}{9}\cdot\underbrace{999\ldots99}_{2016}-\frac{2}{9}\cdot\underbrace{999\ldots99}_{1008}$$ $$=\frac{1}{9}\cdot(10^{2016}-1)-\frac{2}{9}\cdot(10^{1008}-1)$$ $$=\frac{1}{9}\cdot(10^{2016}-1-2\cdot10^{1008}+2)$$ $$=\frac{1}{9}\cdot(10^{2016}-2\cdot10^{1008}+1)$$ $$=\frac{1}{9}\cdot(10^{1008}-1)^2=\left(\frac{10^{1008}-1}{3}\right)^2.$$

Kako je $10^{1008}-1=\underbrace{999\ldots99}_{1008}$ broj djeljiv s 9, pa i s 3, to je $\frac{10^{1008}-1}{3}$ cijeli broj pa je $a-b$ kvadrat cijelog broja.

Drugi način:

$$a-b=\underbrace{111\ldots11}_{2016}-\underbrace{222\ldots22}_{1008}=\underbrace{111\ldots11}_{2016}-2\cdot\underbrace{111\ldots11}_{1008}$$ $$=\underbrace{111\ldots11}_{1008}\cdot\underbrace{100\ldots01}_{1007}-2\cdot\underbrace{111\ldots11}_{1008}$$ $$=\underbrace{111\ldots11}_{1008}\cdot\left(\underbrace{100\ldots01}_{1007}-2\right)=\underbrace{111\ldots11}_{1008}\cdot\underbrace{999\ldots99}_{1008}$$ $$=\underbrace{111\ldots11}_{1008}\cdot9\cdot\underbrace{111\ldots11}_{1008}=\underbrace{111\ldots11}_{1008}^{2}\cdot3^2=\underbrace{333\ldots33}_{1008}^{2}.$$

Time je tvrdnja dokazana.