2. Find all pairs of natural numbers whose difference of squares is 2023.
2. Find all pairs of natural numbers whose difference of squares is 2023.

## Solution.

If $a$ and $b$ are natural numbers, then $a^{2}-b^{2}=2023$.

$a^{2}-b^{2}=(a+b)(a-b)$

Since $a+b > a-b$, we have 3 solutions:*

I. $(a+b)(a-b)=2023 \cdot 1$

$a+b=2023$

$\underline{a-b=1}$

$2a=2024$

$a=1012 \quad b=1011$

II. $(a+b)(a-b)=289 \cdot 7$

$a+b=289$

$\underline{a-b=7}$

$2a=296$

$a=148 \quad b=141$

III. $(a+b)(a-b)=119 \cdot 17$

$a+b=119$

$\underline{a-b=17}$

$2a=136$

$a=68 \quad b=51$

2 POINTS

TOTAL 10 POINTS

*Note. Instead of using $a+b > a-b$, we can examine 3 cases where $a+b > a-b$, or alternatively, not check all 6 possibilities, in which case the student cannot get that point, i.e., can get a maximum of 9 points.