Calculations modulo an integer can sometimes be used to show that an equation has no integer solutions. By working in Z/4Z for example, we can show that a multiple of four plus three cannot be written as a sum of two (integer) squares. In Z/4Z the elements 0 and 1 are precisely the squares of some element: 0² = 0, 1² = 1, 2² = 0, en 3² = 1. If m is an integer, then m ² = 0 or 1 mod 4. Now if m and n are integral, then m² + n² = 0, 1 or 2 mod 4. So m² + n² is not a multiple of four plus three.

Suppose x = (a_n a_n-1 ... a₁ a₀ )₁₀. The well-known `nine test'

9 | x if and only if 9 | a_n + ··· + a₀

is based on modular arithmetic. For this we have to work modulo 9. We have 10 = 1 mod 9, so 10ⁿ = 1ⁿ mod 9 = 1 mod 9. In general we find that the decimal number

x = (a_n a_n-1 ... a₁ a₀ )₁₀ (so x = a_n 10ⁿ + a_{n - 1} 10^{n - 1} + ··· + a₁ 10 + a₀)

modulo 9 equals

x = (a_n a_{n - 1} ... a₁ a₀)₁₀ mod 9

= a_n 10ⁿ + a_{n - 1} 10^{n - 1} + ··· + a₁10 + a₀ mod 9

= a_n + ··· + a₀ mod 9.

In particular:

9 | x if and only if 9 | a_n + ··· + a₀.

Modular arithmetic can greatly reduce the amount of work if we have to compute with powers. For example, we show that 10⁹ + 1 is divisible by 19 as follows. Working modulo 19 we find successively: 10 = -9 mod 19, 10² = 81 = 5 mod 19, 10⁴ = 25 = 6 mod 19, 10⁸ = 36 = -2 mod 19. So 10⁹ = -2 · 10 = -20 = -1 mod 19. But this means that 19 | 10⁹ + 1.