Assume deg P = d * Lemma: We can choose $\pm$ such that $\pm P(x + 1) \pm P(x+2) \pm P(x+3) \pm...... \pm P(x + 2^{d}) = c \ \ \forall x$ just notice that $Q(x+k) - Q(x)$ always decrease degree by one pair them up then repeat So we can increase\decrease the sequence $\pm P(1) \pm P(2) \pm \ldots \pm P(m)$ by c Now we only need to prove that the sequence covers all residue modulo c since $gcd(P(1), \ldots , P(k), \ldots) = 1$ then exist k such that $gcd(P(k),c) = 1$ assume that k =< c By using mod 2, we can show that c is even now for the sequence $A = \pm P(1) \pm P(2) \pm \ldots \pm P(2mc)$, we will choose pair $+P(x+c) - P(x) \ \ \forall x \neq k$ (mod c) and $+P(x) \ \ \forall x = k$ (mod c) $A = (P(c+1) - P(1)) + (P(c+2) - P(2)).... + (P(3c+1) - P(2c + 1)+ .... + 2mP(k) = 2mP(k)$ (mod c) so A covers all even residue mod c We can show that P(1) + P(2) or P(1) odd if P(1) odd choose sign same as above for the sequence $B = \pm P(1) \pm P(2) \pm \ldots \pm P(2mc +1)$ so $B = 2mP(k) + P((2mc + 1) = 2mP(k) + P(1)$ (mod c) which means B covers all odd residue mod c if $P(1) + P(2)$ odd same as above but for the sequence $C = \pm P(1) \pm P(2) \pm \ldots \pm P(2mc +2)$ So we can choose sign such that the sequence covers all residue mod c But we can add/remove c so the sequence covers all integers and each integers can be represented infinitely ways (keep add c and remove c).

Assume deg P = d * Lemma: We can choose $\pm$ such that $\pm P(x + 1) \pm P(x+2) \pm P(x+3) \pm...... \pm P(x + 2^{d}) = c \ \ \forall x$ just notice that $Q(x+k) - Q(x)$ always decrease degree by one pair them up then repeat So we can increase\decrease the sequence $\pm P(1) \pm P(2) \pm \ldots \pm P(m)$ by c Now we only need to prove that the sequence covers all residue modulo c By using mod 2, we can show that c is even since $gcd(P(1), \ldots , P(k), \ldots) = 1$ then exist m such that $gcd(P(1), \ldots , P(m)) = 1$ so by Bezout's theorem, exist $x_1, x_2, \dots x_m$ such that $x_1P(1) + \dots + x_mP(m) = 1$ Let k be the smallest number such that higher than m and c | k now we will choose sign $P(cx + 1) + \dots \pm P(cx + i) + \dots + P(cx + m) \dots + P(cx + h) - P(cx +h + 1) \dots \pm P(cx + i) \dots - P(cx + m + h) \dots - P(cx + 2h) = \pm 2P(i)$ (mod c) so we can pair up 2h numbers to add\remove $2P(i)$ (mod c) for any $1 \leq i \leq m$ So there exist a sequence T such that $T = 2a(x_1P(1) + \dots + x_mP(m)) = 2a$ (mod c) so T covers all even residues mod c We can show that P(1) + P(2) or P(1) odd if P(1) odd choose sign $C = T + P(cy + 1)$ so $B = T + P(cy + 1) = 2a + P(1)$ (mod c) which means B covers all odd residue mod c if $P(1) + P(2)$ odd same as above but for the sequence $C = T + P(cy+1) + P(cy+2)$ So we can choose sign such that the sequence covers all residue mod c But we can add/remove c so the sequence covers all integers and each integers can be represented infinitely ways (keep add c and remove c).