Let see who is smart

Relevant wiki: Euler's Theorem

$\begin{aligned} \sum_{k=0}^{144} (1729+2k)^{1728+2k} & \equiv \sum_{k=0}^{144} (1729+2k)^{\color{#3D99F6}1728+2k \bmod \phi (8)} \text{ (mod 8)} & \small \color{#3D99F6} \text{Since }\gcd(1729+2k,8) = 1 \text{, Euler's theorem applies.} \\ & \equiv \sum_{k=0}^{144} (1729+2k)^{\color{#3D99F6}1728+2k \bmod 4} \text{ (mod 8)} & \small \color{#3D99F6} \text{Euler's totient function }\phi (8) = 4 \\ & \equiv \sum_{k=0}^{144} (1729+2k)^{\begin{cases} 0 & \text{if }k \text{ is even.} \\ 2 & \text{if }k \text{ is odd.} \end{cases}} \text{ (mod 8)} \\ & \equiv \left(\sum_{k=0}^{72} (1729+4k)^0 + \sum_{k=0}^{71} (1731+4k)^2\right) \text{ (mod 8)} \\ & \equiv \left(\sum_{k=0}^{72} 1 + \sum_{k=0}^{71} (1728+3+4k)^2\right) \text{ (mod 8)} \\ & \equiv \left(73 + \sum_{k=0}^{71} (3+4k)^2\right) \text{ (mod 8)} \\ & \equiv \left(1 + \sum_{k=0}^{71} (9+12k+16k^2)\right) \text{ (mod 8)} \\ & \equiv \left(1 + \sum_{k=0}^{71} (1+4k) \right) \text{ (mod 8)} \\ & \equiv \left(1 + 72 +4 \times \frac {71(72)}2 \right) \text{ (mod 8)} \\ & \equiv \left(1 + 0+4 + 0 \right) \text{ (mod 8)} \\ & \equiv \boxed{1} \text{ (mod 8)} \end{aligned}$

Firstly note that odd numbers are raised to even powers which means that each term of the sum is the square of an odd number. Any number of the form $(2n+1)^{2}=4n(n+1)+1$ is congruent to 1 mod 8. Since there are 145 such terms in the sum, the remainder when it is divided by 8, is the same as when 145 is divided by 8. This,very obviously,is 1.