Question

N 3 la puterea 34 ori 4 la puterea 4 la putereav23 - 21 la puterea 21 ori 6 la puterea 23

Answer 1

 n = 3²³ · 4²³ - 2²¹ · 6²³ 

n = 3²³ · (2²)²³ - 2²¹ · (2 · 3)²³=>n = 3²³ · 2²°²³ - 2²¹ · 2²³ · 3²³

n = 3²³ · 2⁴⁶ - 2²¹⁺²³ · 3²³

n = 3²³ · 2⁴⁶ - 2⁴⁴ · 3²³

n = 3²³ · (2⁴⁶ - 2⁴⁴)

n = 3²³ · 2⁴⁴ · (2² - 1)

n = 3²³ · 2⁴⁴ · (4 - 1)

n = 3²³ · 2⁴⁴ · 3

n = 3²³⁺1 · 2⁴⁴ 

n = 3²⁴ · 2⁴⁴

n = (3²)¹² · (2²)²²

3² și 2² sunt pătrate perfecte