Solutions with Integer Values | 整数解问题

📚 Solutions with Integer Values | 整数解问题

In A-Level mathematics, many problems ask for solutions that must be integers, rather than any real number. This constraint arises in number theory, algebra, inequalities, and applied problems, and it often requires careful reasoning beyond simply solving an equation.

在A-Level数学中，许多问题要求解必须是整数，而不是任意实数。这个约束出现在数论、代数、不等式和应用题中，往往需要比单纯求解方程更细致的推理。

1. What Does “Integer Solution” Mean? | 什么是“整数解”

An integer solution is a solution to an equation, inequality, or system where every variable takes an integer value. The set of integers includes positive and negative whole numbers and zero: {…, −3, −2, −1, 0, 1, 2, 3, …}.

整数解是指方程、不等式或方程组中每个变量均取整数值的解。整数集合包括正整数、负整数和零：{…，−3，−2，−1，0，1，2，3，…}。

For example, the equation 2x = 7 has no integer solution, because x = 3.5 is not an integer. However, 2x = 8 has the integer solution x = 4.

例如，方程 2x = 7 没有整数解，因为 x = 3.5 不是整数。但 2x = 8 有整数解 x = 4。

2. Linear Diophantine Equations: ax + by = c | 线性丢番图方程：ax + by = c

A linear Diophantine equation is an equation of the form ax + by = c, where a, b, c are integers and we seek integer solutions for x and y. These equations appear in number theory and in certain real-world problems like exchanging coins.

线性丢番图方程是形如 ax + by = c 的方程，其中 a、b、c 为整数，我们寻找 x 和 y 的整数解。这类方程出现在数论以及诸如换硬币等现实问题中。

The key fact is that such an equation has integer solutions if and only if the greatest common divisor of a and b divides c. In symbols:

关键事实是：这样的方程存在整数解当且仅当 a 和 b 的最大公约数整除 c。用符号表示：

gcd(a, b) | c ⇔ integer solutions exist

If this condition holds, there are infinitely many integer solutions, given by a general formula.

如果该条件成立，则存在无穷多个整数解，并由一个一般公式给出。

3. Solving ax + by = c Using the Euclidean Algorithm | 用欧几里得算法求解 ax + by = c

One method to find one particular integer solution is the extended Euclidean algorithm. First find numbers u and v such that au + bv = gcd(a, b). Then multiply u and v by c / gcd(a, b) to get a particular solution.

求一组特定整数解的一种方法是扩展欧几里得算法。首先找到 u 和 v，使得 au + bv = gcd(a, b)。然后将 u 和 v 乘以 c / gcd(a, b) 得到一组特解。

x₀ = u × (c / d), y₀ = v × (c / d) where d = gcd(a, b)

All integer solutions are then given by:

所有整数解则由下式给出：

x = x₀ + (b/d)t, y = y₀ − (a/d)t, t ∈ ℤ

Here t is any integer parameter.

其中 t 是任意整数参数。

4. Worked Example: 14x + 9y = 1 | 例题：14x + 9y = 1

Let us solve 14x + 9y = 1. First compute gcd(14, 9). Since 14 = 9 × 1 + 5, 9 = 5 × 1 + 4, 5 = 4 × 1 + 1, the gcd is 1. Working backwards:

我们来解 14x + 9y = 1。首先计算 gcd(14, 9)。因为 14 = 9 × 1 + 5，9 = 5 × 1 + 4，5 = 4 × 1 + 1，所以最大公约数为 1。倒推回去：

1 = 5 − 4 × 1 = 5 − (9 − 5) = 2 × 5 − 9 = 2 × (14 − 9) − 9 = 2 × 14 − 3 × 9

Thus a particular solution is x = 2, y = −3. The general solution is:

因此一组特解为 x = 2，y = −3。通解为：

x = 2 + 9t, y = −3 − 14t, t ∈ ℤ

Check: 14(2) + 9(−3) = 28 − 27 = 1. For t = 1 we get x = 11, y = −17, and indeed 14(11) + 9(−17) = 154 − 153 = 1.

验证：14(2) + 9(−3) = 28 − 27 = 1。当 t = 1 时得到 x = 11，y = −17，确实有 14(11) + 9(−17) = 154 − 153 = 1。

5. Systems of Equations with Integer Constraints | 带整数约束的方程组

Sometimes a system of linear equations may have infinitely many real solutions, but only finitely many or even no integer solutions. For example, consider:

有时一个线性方程组可能有无数实数解，但只有有限个甚至没有整数解。例如，考虑：

x + y = 5, x − y = 1

Solving gives x = 3, y = 2, which are integers. But if the equations were x + y = 5 and x − y = 0.2, the solution would be x = 2.6, y = 2.4, not integers.

解得 x = 3，y = 2，均为整数。但如果方程组为 x + y = 5 和 x − y = 0.2，则解为 x = 2.6，y = 2.4，不是整数。

For a system to have integer solutions, the coefficients and constants must satisfy certain divisibility conditions after elimination.

一个方程组要有整数解，消元后系数与常数必须满足一定的整除条件。

6. Quadratic Equations with Integer Roots | 有整数根的二次方程

If a quadratic equation with integer coefficients has integer roots, those roots must divide the constant term. This is a direct consequence of the factor theorem and Vieta’s formulas.

如果整数系数的二次方程有整数根，那么这些根必定整除常数项。这是因式定理和韦达定理的直接推论。

For a quadratic x² + bx + c = 0 with integer roots r and s, we have:

对于具有整数根 r 和 s 的二次方程 x² + bx + c = 0，有：

r + s = −b, r × s = c

Therefore, to find integer roots, test all integer factor pairs of c.

因此，要寻找整数根，需要检验 c 的所有整数因子对。

Example: x² − 5x + 6 = 0. The factor pairs of 6 are (1,6), (−1,−6), (2,3), (−2,−3). The pair (2,3) sums to 5, so the roots are 2 and 3.

例：x² − 5x + 6 = 0。6 的因子对有 (1,6)、(−1,−6)、(2,3)、(−2,−3)。其中 (2,3) 的和为 5，所以根是 2 和 3。

7. Inequalities and Integer Solutions | 不等式与整数解

When an inequality has infinitely many real solutions, its integer solutions are often listed as a set of consecutive or discrete integers. For example, solve −3 < 2x − 1 < 7.

当一个不等式有无数实数解时，它的整数解通常是一组连续或不连续的整数。例如，解 −3 < 2x − 1 < 7。

Add 1 to all parts: −2 < 2x < 8, then divide by 2: −1 < x < 4. The integer solutions are x = 0, 1, 2, 3.

各部分加 1：−2 < 2x < 8，然后除以 2：−1 < x < 4。整数解为 x = 0，1，2，3。

For quadratic inequalities like x² − 4x + 3 ≤ 0, factor to get (x − 1)(x − 3) ≤ 0. The real solution interval is 1 ≤ x ≤ 3, so the integer solutions are x = 1, 2, 3.

对于二次不等式如 x² − 4x + 3 ≤ 0，因式分解得 (x − 1)(x − 3) ≤ 0。实数解区间为 1 ≤ x ≤ 3，所以整数解为 x = 1，2，3。

8. Integer Solutions in Exponential and Mod Equations | 指数与模方程中的整数解

Equations involving powers or absolute values may have integer solutions that can be found by reasoning about bounds. For instance, solve 2ⁿ = 32. Since 32 = 2⁵, the integer solution is n = 5.

涉及幂或绝对值的方程可以通过有界推理来寻找整数解。例如，解 2ⁿ = 32。因为 32 = 2⁵，所以整数解为 n = 5。

For |x − 2| = 5, the solutions are x − 2 = 5 or x − 2 = −5, giving x = 7 or x = −3, both integers.

对于 |x − 2| = 5，解为 x − 2 = 5 或 x − 2 = −5，得到 x = 7 或 x = −3，均为整数。

In modular arithmetic, an equation like 3x ≡ 4 (mod 7) asks for integers x such that 3x − 4 is a multiple of 7. Testing small residues gives x ≡ 6 (mod 7), so x = 6 + 7t.

在模算术中，方程如 3x ≡ 4 (mod 7) 要求找到整数 x 使得 3x − 4 是 7 的倍数。检验小的余数得到 x ≡ 6 (mod 7)，所以 x = 6 + 7t。

9. Applied Problems: Coins, Tickets, and Mixing | 应用题：硬币、门票与混合问题

A classic application is: “A shopkeeper has only £3 coins and £5 coins. Can he pay exactly £44?” This is equivalent to solving 3x + 5y = 44 in non-negative integers.

一个经典应用是：“店主只有 3 英镑硬币和 5 英镑硬币。他能正好支付 44 英镑吗？”这等价于求 3x + 5y = 44 的非负整数解。

Since gcd(3, 5) = 1 divides 44, integer solutions exist. One solution is x = 8, y = 4, because 3(8) + 5(4) = 24 + 20 = 44. Other solutions with non-negative values include x = 3, y = 7 (9 + 35 = 44) and x = 13, y = 1 (39 + 5 = 44).

因为 gcd(3, 5) = 1 整除 44，所以整数解存在。一个解是 x = 8，y = 4，因为 3(8) + 5(4) = 24 + 20 = 44。其他非负解包括 x = 3，y = 7（9 + 35 = 44）和 x = 13，y = 1（39 + 5 = 44）。

In such problems, we must also apply non-negativity constraints, which restrict the parameter t to a finite range.

在这类问题中，我们还必须应用非负约束，这会限制参数 t 在一个有限范围内。

10. Common Mistakes and Tips | 常见错误与技巧

A common mistake is forgetting to check whether a solution is an integer. For example, solving 6x + 2 = 17 gives x = 2.5, which is not an integer. In integer solution problems, always state whether the variable is allowed to be negative or zero.

一个常见错误是忘记检查解是否为整数。例如，解 6x + 2 = 17 得到 x = 2.5，这不是整数。在整数解问题中，始终要说明变量是否允许为负数或零。

Another error is using real-number reasoning when a divisibility test is required. For instance, to find integer solutions to 10x + 15y = 7, note that gcd(10, 15) = 5, but 5 does not divide 7, so no integer solutions exist.

另一个错误是当需要整除检验时却使用实数推理。例如，求 10x + 15y = 7 的整数解，注意 gcd(10, 15) = 5，但 5 不能整除 7，所以不存在整数解。

Useful tips: always factor or simplify first, test factor pairs, and use modular arithmetic when numbers are large.

有用技巧：始终先因式分解或化简，检验因子对，并在数字较大时使用模算术。

11. Practice Problems | 练习题目

Try the following problems to strengthen your understanding:

尝试以下题目以巩固理解：

1. Find all integer solutions to 6x + 8y = 10.
2. Find all integer solutions to x² − 7x + 10 = 0.
3. Solve the inequality 3 ≤ 2x − 1 ≤ 7 and list the integer solutions.
4. Determine whether 4x + 6y = 9 has any integer solutions.

1. 求 6x + 8y = 10 的所有整数解。
2. 求 x² − 7x + 10 = 0 的所有整数解。
3. 解不等式 3 ≤ 2x − 1 ≤ 7 并列出整数解。
4. 判断 4x + 6y = 9 是否存在整数解。

Answers: 1. gcd(6,8)=2, divide by 2: 3x+4y=5, one solution x=3, y=−1, general x=3+4t, y=−1−3t. 2. x=2 or x=5. 3. 2 ≤ x ≤ 4, integers 2,3,4. 4. gcd(4,6)=2 does not divide 9, so no solutions.

答案：1. gcd(6,8)=2，两边除以 2：3x+4y=5，一个特解 x=3，y=−1，通解 x=3+4t，y=−1−3t。2. x=2 或 x=5。3. 2 ≤ x ≤ 4，整数为 2，3，4。4. gcd(4,6)=2 不整除 9，所以无解。

12. Summary | 总结

Integer solution problems test your ability to combine algebraic manipulation with number-theoretic conditions. The key steps are: identify the type of equation or inequality, determine any divisibility or parity conditions, and then systematically list or parameterize the solutions.

整数解问题考查你将代数操作与数论条件相结合的能力。关键步骤是：识别方程或不等式的类型，确定整除性或奇偶性条件，然后系统地列出或用参数表示解。

For linear Diophantine equations, always check the gcd condition and use the extended Euclidean algorithm for particular solutions. For quadratics, factor and use the factor-pair method. For inequalities, find the real interval and select the integers within it.

对于线性丢番图方程，始终检查最大公约数条件，并使用扩展欧几里得算法求特解。对于二次方程，因式分解并使用因子对方法。对于不等式，求出实数区间并选择其中的整数。

With enough practice, these problems become straightforward. Always verify your answers by substituting back.

通过足够练习，这类问题会变得简单直接。始终通过代入回验来验证答案。

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Solutions with Integer Values | 整数解问题