Permutation of the Year 1

We have $2026=45^2+1$. Therefore, for every integer $d$,

$$d^2+(45d)^2=2026d^2$$

is divisible by $2026$.

Define

$$a_i=(45(i-1)\bmod 2026)+1.$$

Since $\gcd(45,2026)=1$, the values $a_i$ for $i=1,2,\cdots,2026$ are exactly $1,2,\cdots,2026$ in some order.

For every $1 \leq i < j \leq 2026$,

$$a_i-a_j \equiv 45(i-j) \pmod{2026}.$$

Thus

$$(i-j)^2+(a_i-a_j)^2 \equiv (1+45^2)(i-j)^2 \equiv 0 \pmod{2026}.$$

So the constructed permutation is valid.

Solution written by GPT5.5