Let $n_{1}, n_{2}, ... n_{k} $ be a sequence of k consecutive odd integers. If $n_{1} + n_{2} + n_{3} = p^3$ and $n_{k} + n_{k-1} + n_{k-2} + n_{k-3} + n_{k-4} = q^4$ where both p and q are prime, what is k?
I am struggling with this question. I know the first sum can be written as $3n_{1} + 6 = p^3$ and the second sum can be written as $5n_{k} - 20 = q^4$. I believe the second sum is also $5n_{1} +10k - 30 = q^4$. However rearranging these I get no workable equations.
Hints:
The sum of $3$ consecutive odd integers is always a multiple of $3$.
What does that imply about $p$?
Can you then find $n_1,n_2,n_3$?
The sum of $4$ odd numbers is always a multiple of $2$.
What does that imply about $q$?
Can you finish?
Note:$\;$The wording of the problem is a little sneaky. The phrase "consecutive odd integers" doesn't automatically mean consecutive increasing (although by default, it usually does). Keep that in mind.
It is a little appreciated fact that every power $k\ge 2$ of any positive integer $n$ can be expressed as the sum of exactly $n$ consecutive odd numbers, viz: $$n^k=\sum_{i=\frac{n^{k-1}-n}{2}+1}^{\frac{n^{k-1}+n}{2}}(2i-1)$$ So $3$ consecutive odd numbers can be found that sum to $3^3$ and $5$ consecutive odd numbers can be found that sum to $5^4$. Using the formula above, it is easy to find that the smallest of the $3$ numbers is $2\cdot(\frac{3^{2}-3}{2}+1)-1=7$ and the largest of the $5$ numbers is $2\cdot(\frac{5^{3}+5}{2})-1=129$
These results conform to your requirement that $p,q$ be prime, as $7+9+11=27=3^3 \Rightarrow p=3$ and $121+123+125+127+129=625=5^4 \Rightarrow q=5$.
To answer your specific question, $2t-1=129 \Rightarrow t=65$ and $2s-1=7 \Rightarrow s=4$. You want to find how many odd numbers are in the sequence beginning at the $4$th odd number and ending at the $65$th odd number. Therefore, the value of $k$ you seek is $65-3=62$.
You got it to $3n_1+6=p^3$ but how about $3(n_1+2)=p p^2$ and $p$ is prime so $p=3$. Using this you can find $n_1$ then $q$ and after all this $k$.
The idea is to figure out $p$, then $n_1$, then $q$ and finally $k$.
$3n_1 + 6 = 3(n_1 + 2) = p^3$ implies that $p$ divides $3(n_1 + 2)$. Since $p$ is prime, then either $p = 3$ or $p \mid (n_1 + 2)$. If $p \neq 3$, then we can run a descent argument to get a contradiction. An explicit run of this argument is given below.
Explicitly: if $p \neq 3$, then $p \mid (n_1+2)$, which implies that $n_1 + 2 = p \ell$ for some $\ell \in \mathbb{Z}$. Substituting in the original equation, we get $3p\ell = p^3$ or $3 \ell = p^2$. But $p$ is prime and is not $3$, so $p \mid \ell$ and $\ell = p \ell'$. Substituting, we get $3p^2 \ell' = p^3$ or $3\ell' = p$. Again, $p$ is prime but is not $3$ so we can write $\ell' = p \ell''$, and substituting yields $3p^3 \ell'' = p^3$ or $3 \ell'' = 1$ which is a contradiction as $\ell''$ is an integer and $3$ is not a unit in $\mathbb{Z}$.
Thus, $p = 3$ and so $n_1 = 7$.
How about $q$? As you suggested, we have $5(n_1 + 2k - 6) = q^4$. Using $n_1 = 7$, we get $5(1 + 2k) = q^4$. Since $q$ is prime, then either $q = 5$ or $q \mid (1 + 2k)$. If $q \neq 5$, we can run a descent argument again to arrive at a contradiction.
So $q = 5$ and $q^4 = 625$. This implies that $k = 62$.