x² − y² = 2024

Idea. A product of two integers with prescribed sum parity: enumerate factor pairs.

Parity. $x-y$ and $x+y$ have the same parity. Their product $2024=2^3\cdot11\cdot23$ is even, so both factors are even. Write $x-y=2u$, $x+y=2v$ with $uv=506=2\cdot11\cdot23$. Then $x=u+v$, $y=v-u$.

Positive factor pairs of $506$. Divisors: $1,2,11,22,23,46,253,506$. With $u\le v$ and $uv=506$:
$$(1,506),\ (2,253),\ (11,46),\ (22,23).$$

Non-negative solutions $(x,y)=(u+v,v-u)$.
$$(507,505),\ (255,251),\ (57,35),\ (45,1).$$

All integer solutions. Changing the signs of $x$ and/or $y$ gives $4\cdot4=16$ solutions: $(\pm507,\pm505)$, $(\pm255,\pm251)$, $(\pm57,\pm35)$, $(\pm45,\pm1)$.

Check. $45^2-1=2024$ ✓.

Remark. A number $N$ is a difference of two squares iff $N\not\equiv2\pmod4$; $2024\equiv0\pmod4$ works.

Lesson. Factor the equation into a product of two integers, then enumerate divisors using parity and size constraints.