Squarey Sum

It can be found that $A_{k}=\frac{k(k+1)(2k+1)(3k^{2}+3k-1)}{30}$ by solving the recurrence relation with $A_{1}=1$ and $A_{k}-A_{k-1}=k^{4}$ (note: assume $A_{k}$ is in the form $a_{1}x^{5}+a_{2}x^{4}+a_{3}x^{3}+a_{4}x^{2}+a_{5}x+a_{6}$ and solve for the coefficients).

We also know that $B_{k}=\frac{k(k+1)(2k+1)}{6}$ and $C_{k}=\frac{k(k+1)}{2}$ .

$\frac{A_{k}}{B_{k}}=\frac{6(3k^{2}+3k-1)}{30}=\frac{3k^{2}+3k-1}{5}$ . This value is an integer when $k≡1$ or $3\mod5$ .

$\frac{B_{k}}{C_{k}}=\frac{2(2k+1)}{6}=\frac{2k+1}{3}$ . This value is an integer when $k≡1\mod3$ .

Therefore, solving this system of linear congruences, we have $k≡1$ or $13\mod15$ .

The values of $1≤k≤100$ that satisfy the above linear congruence are $1,13,16,28,31,43,46,58,61,73,76,88,91$ . These values sum up to $\boxed{625}$ .

Note: I have skipped a few steps in the solution. If you would like me to expand on a part of it, feel free to ask me.

Direct computation confirms Sam Zhou's solution:

$a(\text{k\_})\text{:=}\sum _{i=1}^k i^4,\,b(\text{k\_})\text{:=}\sum _{i=1}^k i^2,\,c(\text{k\_})\text{:=}\sum _{i=1}^k i$

$\text{Table}[\alpha =a(j);\beta =b(j);\gamma =c(j);\text{If}[(\alpha \bmod \beta )=0\land (\beta \bmod \gamma )=0,j,\text{Nothing}],\{j,100\}] \Rightarrow \{1,13,16,28,31,43,46,58,61,73,76,88,91\}$

Which totals to $625$ or $25^2$ .