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