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.

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

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

(5)

题目中文翻译

(i) 某学生写道：“当 $a$ 和 $b$ 是连续素数时， $a^2+b^2$ 绝不会是10的倍数。“用反例证明该命题不成立。

(ii) 已知 $x$ 和 $y$ 是大于0且小于6的偶数，用穷举法证明 $1 < x^2 - \dfrac{xy}{4} < 15$ 。

（5分）

解答

(i) 找两个连续素数 $a,b$ 使得 $a^2+b^2$ 是10的倍数，例如 $7^2+11^2=170$ 。(ii) 偶数 $x,y\in\{2,4\}$ ，共4种组合，逐一计算验证。

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

$a^2 + b^2 = 49 + 121 = 170 = 10 \times 17$

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

(ii) 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 $(x, y)$ . $\quad\blacksquare$