sufficient condition of identical congruence

Theorem.  Let  $f(X):=a_n{X}^n+\mathrm{\dots }+a_{1}X+a_0$  be a polynomial in $X$ with integer coefficients $a_i$ and $m$ a positive integer.  If the congruence

$f(x)\equiv \mathrm{\hspace{0.33em}0}(modm)$(1)

is satisfied by $m$ successive integers $x$, then it is satisfied by all integers $x$, in other words it is an identical congruence.

Proof.  There is an integer $x_0$ such that (1) is satisfied by

$$x:=x_0+1,x_0+2,\mathrm{\dots },x_0+m.$$

But these values form a complete residue system modulo $m$.  Thus, if $x$ is an arbitrary integer, one has

$$x\equiv x_0+r(modm)\mathit{\hspace{1em}}\text{where}\mathrm{\hspace{0.33em}\hspace{0.33em}1}\leqq r\leqq m.$$

This implies

$$a_i{x}^i\equiv a_i{(x_0+r)}^i(modm)\mathit{\hspace{1em}}\text{for}i=0,\mathrm{\hspace{0.17em}1},\mathrm{\dots },n$$

and consequently

$$\underset{f(x)}{\underbrace{\sum _{i=0}^n{a}_i{x}^i}}\equiv \sum _{i=0}^n{a}_i{(x_0+r)}^i=f(x_0+r)\equiv \mathrm{\hspace{0.33em}0}(modm).$$

Accordingly, (1) is true for any integer $x$, Q.E.D.

Note.  Though the congruence (1) is identical, it need not be a question of a formal congruence

$f(X)\underset{¯}{\equiv }\mathrm{\hspace{0.33em}0}(modm),$(2)

i.e. all coefficients $a_i$ need not be congruent to 0 modulo $m$.