2024/020) Show that there are infinitely many composite numbers of the form $10^n +3$ (n = 1, 2, 3, ... ).

If n = 2 we have $10^2+3 = 103$ this is divisible by 103 . 

Hence  $10^2 + 3 = 0 \pmod {103}\cdots(1)$

We know that 103 is a prime but we shall use this as a starting point to show that there are infinite numbers of the form $10^n+3$ is divisible by 103.

As 103 is a prime so using Fermats little theorem

$10^{102} = - 1 \pmod {103}$

Squaring both sides we get

$10^{204} =  1 \pmod {103}$

So from (1) and above we have

$10^{204n + 2} + 3=  0 \pmod {103}$

except for n = 0 when the value is 103 all other values are above 103 and have a factor 103 so not a prime

So there are infinitely many composite numbers as they are divisible by 103. We have not taken all cases and have found infinitely many composite numbers. Hence proved