题目

Problem

(i) A student writes the following statement:

“When $a$ and $b$ are consecutive prime numbers, $a^2 + b^2$ is never a multiple of 10”

Prove by counter example that this statement is not true.

(2)

(ii) Given that $x$ and $y$ are even integers greater than 0 and less than 6, prove by exhaustion, that

1 < x^2 - \frac{xy}{4} < 15

(3)

题目中文翻译

(i) 一名学生写下命题：“当 $a,b$ 是相邻的素数时， $a^2+b^2$ 从不是 10 的倍数。”用反例证明这个命题不成立。（2 分）

(ii) 已知 $x,y$ 是大于 0、小于 6 的偶整数，用穷举法证明

1<x^2-\frac{xy}{4}<15.

（3 分）

解答

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选两个连续素数，使平方和恰为 10 的倍数；一个反例即可推翻原命题。

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Let $a = 7$ and $b = 11$ , which are consecutive prime numbers.

\begin{align*} a^2 + b^2 =&\, 49 + 121 \\ =&\, 170 \\ =&\, 10 \times 17 \end{align*}

Since $170$ is a multiple of $10$ , the statement is not true. $\quad\blacksquare$

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$x,y$ 只能取 2 或 4，共四个有序组合。逐一计算，确认所有值均严格介于 1 和 15 之间。

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The even integers satisfying $0 < x, y < 6$ are $x, y \in \{2, 4\}$ .

The four possible combinations:

$x$	$y$	$x^2$	$\dfrac{xy}{4}$	$x^2 - \dfrac{xy}{4}$
$2$	$2$	$4$	$1$	$3$
$2$	$4$	$4$	$2$	$2$
$4$	$2$	$16$	$2$	$14$
$4$	$4$	$16$	$4$	$12$

For all four combinations:

1 < x^2 - \frac{xy}{4} < 15

The inequality holds for all valid $(x, y)$ . $\quad\blacksquare$

也可将不等式乘以正数 4，检查 $4<4x^2-xy<60$ 。按表中相同顺序，四个结果为 12、8、56、48，全部满足，且覆盖所有组合。