A riddle - is there a way to solve it non-numerically?

Question

So the riddle is:

John has written down $k$ sequential odd numbers: $n_1, n_2, ..., n_{k-1}, n_{k}$ (where $n_{2} = n_{1} + 2$ and so on). We know that:

• The sum of the first four numbers is a fourth power of some prime number (so $n_{1} + n_{2} + n_{3} + n_{4} = p_{1}^4$ where $p_{1}$ is a prime number.
• The sum of the last five numbers is a fourth power of some prime number (so $n_{k} + n_{k-1} + n_{k-2} + n_{k-3} + n_{k-4}= p_{2}^4$ where $p_{2}$ is a prime number.

The question is - how many numbers have been written down ($k=?$).

I've tried to solve it numerically, but without success - is there a way to solve it analitically?

EDIT: Sorry it's not even, it is ODD!

Answer 1

My guess is that the correct formulation of the problem is:

$n_{1}$ is arbitrary, and then $n_{i+1} = n_{i} + 2$.

In this case, the solution, as indicated in the other answers, should be $$ 1,3,5,7,\dots,121,123,125,127,129. $$

Proof.

The sum of four consecutive such numbers is $4$ times the average of the first and last one, so for the first question the average must be $4$ (so that you get the fourth power of $2$), and the four numbers are $1,3,5,7$.

The sum of five consecutive such numbers is $5$ times the average of the first and last one, so for the second question the average must be $125$ (so that you get the fourth power of $5$), and the five numbers are $121,123,125,127,129$.

Answer 2

If the sum of the first four is the fourth power of some prime then (assuming the first number is $n$) we get:$$n+(n+2)+(n+4)+(n+6)=p^4$$$$4n+12=p^4$$$$4(n+3)=p^4$$This implies $p$ must be even - the only even prime is $2$. Hence:$$4(n+3)=2^4$$Which leads to $n=1$. However, I just noticed in your question that you said the numbers are all even - which means there is no solution.

Answer 3

Well, let $s = n\{1\}$, then $\sum_{i=1}^4 n\{i\}=p^4$ for some $p$, but the lhs is equal to $(s + s+2 + s+4 + s+6) = 8s + 12 = 4(s+3)$, so $4$ divides $p^4$, but $p$ is prime, therefore $p$ must be $2$ and $s + 3$ must be $4$, so $s = 1$.

Now the other piece.

$n\{k\} = 1 + 2k-2$, so $(1+2k-2+1+2k-4+1+2k-6+1+2k-8+1+2k-10) = 10k-25 = 5(2k-5)$ so the other prime is 5, and $2k-5=5^3=125$ so $2k=130$ and $k=65$