To solve the problem, we need to find an integral root \( \alpha \) of the polynomial \( P(x) \) given that \( P(2) = 13 \) and \( P(10) = 5 \).
### Step-by-Step Solution:
1. **Understanding the Remainder Theorem**:
Since \( P(x) \) is a polynomial with integer coefficients and has at least one integral root \( \alpha \), we can use the property that for any integers \( a \) and \( b \):
\[
P(a) - P(b) = P(a - b) \text{ is divisible by } a - b.
\]
2. **Applying the Remainder Theorem**:
Let's apply this theorem to our specific values:
- Let \( a = 2 \) and \( b = \alpha \):
\[
P(2) - P(\alpha) = 13 - 0 = 13 \text{ is divisible by } 2 - \alpha.
\]
Thus, \( 2 - \alpha \) divides \( 13 \).
3. **Finding Divisors of 13**:
The divisors of 13 are \( \pm 1, \pm 13 \). Hence, we have:
- \( 2 - \alpha = 1 \) implies \( \alpha = 1 \)
- \( 2 - \alpha = -1 \) implies \( \alpha = 3 \)
- \( 2 - \alpha = 13 \) implies \( \alpha = -11 \)
- \( 2 - \alpha = -13 \) implies \( \alpha = 15 \)
So, the possible values for \( \alpha \) from this equation are \( 1, 3, -11, 15 \).
4. **Using the Second Condition**:
Now, let’s apply the same reasoning to \( P(10) \):
- Let \( b = \alpha \):
\[
P(10) - P(\alpha) = 5 - 0 = 5 \text{ is divisible by } 10 - \alpha.
\]
Thus, \( 10 - \alpha \) divides \( 5 \).
5. **Finding Divisors of 5**:
The divisors of 5 are \( \pm 1, \pm 5 \). Hence, we have:
- \( 10 - \alpha = 1 \) implies \( \alpha = 9 \)
- \( 10 - \alpha = -1 \) implies \( \alpha = 11 \)
- \( 10 - \alpha = 5 \) implies \( \alpha = 5 \)
- \( 10 - \alpha = -5 \) implies \( \alpha = 15 \)
So, the possible values for \( \alpha \) from this equation are \( 9, 11, 5, 15 \).
6. **Finding Common Values**:
Now we need to find the common integral values of \( \alpha \) from both sets:
- From the first condition: \( 1, 3, -11, 15 \)
- From the second condition: \( 9, 11, 5, 15 \)
The common value is \( \alpha = 15 \).
### Conclusion:
Thus, the integral value of \( \alpha \) that must satisfy \( P(x) = 0 \) is:
\[
\boxed{15}
\]
