A number theory problem

$\begin{aligned} n^3+3n^2+5n+3 & = n^3+3n^2+3n+1 + 2n + 2 \\ & = (n+1)^3 +2(n+1) \\ & = (n+1)(n^2+2n+1+2) \\ & = (n+1)(n[n+2] + 3) \\ & = n(n+1)(n+2) + 3(n+1) \end{aligned}$

For integer $n$ , the term $n(n+1)(n+2)$ is always divisible by $3$ and the term $3(n+1)$ is, of course, divisible by $3$ .

$\Rightarrow n^3+3n^2+5n+3 \equiv \boxed{0} \pmod{3}$

$\color{#3D99F6}{\text{Easier Solution}}$

There is an easier solution though not as elegant.

Since $f(n) = n^3+3n^2+5n+3$ gives the same remainder when divided by $3$ for all $n$ . We can get the solution just be checking for $f(n) \mod{3}$ for any $n$ , says $0$ , then $f(0)=3$ , therefore the remainder is $0$ . Of course, we are assuming the question is correct, to check that we can try more $n$ . $f(1) = 12$ , $f(2) = 33$ .... All are divisible by $3$ .

$n^{3} + 3n^{2} + 5n + 3 = (n^{3} + 3n^{2} + 2n) + (3n + 3) =$

$n(n^{2} + 3n + 2) + 3(n + 1) = n(n + 1)(n + 2) + 3(n + 1).$

Now given any $3$ consecutive integers, exactly one of them will be divisible by $3,$ i.e., exactly one of $n, (n + 1), (n + 2)$ is divisible by $3,$ and thus the product $n(n + 1)(n + 2)$ is divisible by $3$ as well. Since $3(n + 1)$ is also divisible by $3$ for any $n,$ we see that the given polynomial is divisible by $3$ for any integer $n,$ leaving a remainder of $\boxed{0}.$

$n^3 + 3n^2 + 5n + 3$ = $n^3 + 3n^2 + (6n - n) + 3$

= $(n^3 - n)$ + $3(n^2 + 2n + 1)$

= $n(n^2 - 1)$ + $3(n^2 + 2n + 1)$

= $(n - 1)(n)(n + 1)$ + $3(n^2 + 2n + 1)$

Not that any number of the form $(n - 1)(n)(n + 1)$ is divisible by 3 and also $3(n^2 + 2n + 1)$ is always divisible by $3$ for any $n$ .

Therefore the remainder left is $\boxed{0}$

Take n as 1. 1+3+5+3=12. 12 divided by 3 is 4 R0 QED.