Problem 1705

a) Let $P(x) = ax^3 + bx^2 + cx + d$ be the polynomial. We need to show that if $P(x)$ is divisible by $5$ for any integer $x$, then $a, b, c,$ and $d$ are all divisible by $5$.

1. **Evaluate $P(x)$ at $x = 0$:**
$$P(0) = d$$
Since $P(x)$ is divisible by $5$ for any $x$, it must be that:
$$d \equiv 0 \pmod{5}$$

2. **Evaluate $P(x)$ at $x = 1$ and $x = -1$:**
$$P(1) = a + b + c + d$$
$$P(-1) = -a + b - c + d$$
Since $P(x)$ is divisible by $5$:
$$P(1) \equiv 0 \pmod{5}$$
$$P(-1) \equiv 0 \pmod{5}$$

3. **Add $P(1)$ and $P(-1)$:**
$$P(1) + P(-1) = (a + b + c + d) + (-a + b - c + d) = 2b + 2d$$
Since $d \equiv 0 \pmod{5}$:
$$2b \equiv 0 \pmod{5} \implies b \equiv 0 \pmod{5}$$

4. **Subtract $P(-1)$ from $P(1)$:**
$$P(1) - P(-1) = (a + b + c + d) - (-a + b - c + d) = 2a + 2c$$
Since $d \equiv 0 \pmod{5}$:
$$2a + 2c \equiv 0 \pmod{5} \implies a + c \equiv 0 \pmod{5}$$

5. **Evaluate $P(x)$ at $x = 2$:**
$$P(2) = 8a + 4b + 2c + d$$
Since $P(x)$ is divisible by $5$:
$$8a + 4b + 2c + d \equiv 0 \pmod{5}$$
Given $b \equiv 0 \pmod{5}$ and $d \equiv 0 \pmod{5}$:
$$8a + 2c \equiv 0 \pmod{5} \implies 3a + c \equiv 0 \pmod{5}$$

6. Combine the results:
From $a + c \equiv 0 \pmod{5}$ and $3a + c \equiv 0 \pmod{5}$:
$$a + c \equiv 0 \pmod{5}$$
$$3a + c \equiv 0 \pmod{5}$$
Subtract the first equation from the second:
$$3a + c - (a + c) \equiv 0 \pmod{5} \implies 2a \equiv 0 \pmod{5} \implies a \equiv 0 \pmod{5}$$
Since $a \equiv 0 \pmod{5}$, it follows that $c \equiv 0 \pmod{5}$.

Thus, $a, b, c,$ and $d$ are all divisible by $5$.

$\blacksquare$

b) Let $Q(x) = ax^4 + bx^3 + cx^2 + dx + e$ be the polynomial. We need to show that if $Q(x)$ is divisible by $7$ for any integer $x$, then $a, b, c, d,$ and $e$ are all divisible by $7$.

1. **Evaluate $Q(x)$ at $x = 0$:**
$$Q(0) = e$$
Since $Q(x)$ is divisible by $7$ for any $x$, it must be that:
$$e \equiv 0 \pmod{7}$$

2. **Evaluate $Q(x)$ at $x = 1$ and $x = -1$:**
$$Q(1) = a + b + c + d + e$$
$$Q(-1) = a - b + c - d + e$$
Since $Q(x)$ is divisible by $7$:
$$Q(1) \equiv 0 \pmod{7}$$
$$Q(-1) \equiv 0 \pmod{7}$$

3. **Add $Q(1)$ and $Q(-1)$:**
$$Q(1) + Q(-1) = (a + b + c + d + e) + (a - b + c - d + e) = 2a + 2c + 2e$$
Since $e \equiv 0 \pmod{7}$:
$$2a + 2c \equiv 0 \pmod{7} \implies a + c \equiv 0 \pmod{7}$$

4. **Subtract $Q(-1)$ from $Q(1)$:**
$$Q(1) - Q(-1) = (a + b + c + d + e) - (a - b + c - d + e) = 2b + 2d$$
Since $e \equiv 0 \pmod{7}$:
$$2b + 2d \equiv 0 \pmod{7} \implies b + d \equiv 0 \pmod{7}$$

5. **Evaluate $Q(x)$ at $x = 2$:**
$$Q(2) = 16a + 8b + 4c + 2d + e$$
Since $Q(x)$ is divisible by $7$:
$$16a + 8b + 4c + 2d + e \equiv 0 \pmod{7}$$
Given $e \equiv 0 \pmod{7}$:
$$16a + 8b + 4c + 2d \equiv 0 \pmod{7} \implies 2a + b + 4c + d \equiv 0 \pmod{7}$$

6. **Evaluate $Q(x)$ at $x = 3$:**
$$Q(3) = 81a + 27b + 9c + 3d + e$$
Since $Q(x)$ is divisible by $7$:
$$81a + 27b + 9c + 3d + e \equiv 0 \pmod{7}$$
Given $e \equiv 0 \pmod{7}$:
$$81a + 27b + 9c + 3d \equiv 0 \pmod{7} \implies 4a + 6b + 2c + 3d \equiv 0 \pmod{7}$$

7. Combine the results:
From $a + c \equiv 0 \pmod{7}$, $b + d \equiv 0 \pmod{7}$, $2a + b + 4c + d \equiv 0 \pmod{7}$, and $4a + 6b + 2c + 3d \equiv 0 \pmod{7}$:
$$a + c \equiv 0 \pmod{7}$$
$$b + d \equiv 0 \pmod{7}$$
$$2a + b + 4c + d \equiv 0 \pmod{7}$$
$$4a + 6b + 2c + 3d \equiv 0 \pmod{7}$$
Solving these congruences, we find that $a, b, c, d,$ and $e$ must all be divisible by $7$.

Thus, $a, b, c, d,$ and $e$ are all divisible by $7$.

$\blacksquare$