Question pertaining to perfect powers in sums of consecutive integers

Question

I came across this question in the book Challenge and Thrill of Pre-college mathematics:

Prove that there are infinitely many sets of five consecutive integers $a,b,c,d,e$ such that $a+b+c+d+e$ is a perfect cube, and $b+c+d$ is a perfect square.

I took the numbers to be $(n-2),(n-1),n,(n+1),(n+2)$, and then it was not very difficult to find that what was required to prove was the existence of infinite numbers $n$ such that $5n$ and $3n$ are perfect cubes and squares respectively.

After playing around with prime factorizations for a while, I discovered that $675$ was such a number, followed by $3^9\cdot5^2$. Clearly this has something to do with the prime factorization of these integers.

Futhermore, I think it is correct to assume that that multiplying $675$ with any number $p^k$ where $p$ is a prime and where $k\mid3$ and $k\mid2$, such as multiplying by $2^{6}$, which gives us $43200$, which is indeed such a number.

My problem is that I have no idea how I am supposed to write such a proof formally, or even if this is a proof. While I do believe that I have generated an infinite number of positive integers which obey the question, how do I explain it? What am I missing?

Answer 1

You have to find positive integers $n\ge 3$, such that $3n$ is a perfect square and $5n$ is a perfect cube.

Consider $$n=3^a\cdot 5^b$$ with positive integers $a$ and $b$. $3n$ is a perfect square, if $a$ is odd and $b$ is even. $5n$ is a perfect cube, if $a$ is divisible by $3$ and $b$ has the form $3k+2$. Hence, it is enough that $a$ is of the form $6s+3$ and $b$ of the form $6t+2$ , giving infinite many solutions.

Answer 2

You already have it all for demonstrating there are infinite such numbers.

$$n=675\cdot p^6$$ for primes $p>5$ already gives you infinitely many solutions.

Answer 3

You basically want infinitely many $n$ with $5n$ a perfect cube, and $3n$ a perfect square. Indeed, your last argument is enough to produce infinitely such $n$, since $n = 675p^{6l}$, for any prime(in fact any number) $p$ and any $l$ satisfies the given conditions. This is easy to see because : $$ 5n = 5 \times 675 p^{6l} = 5^33^3p^{6l} = (15p^{2l})^3 \quad ; \quad 3n = 3 \times 675 p^{6l} = 5^23^4p^{6l} = (45p^{3l})^2 $$

Since both $p$ and $l$ may infinitely vary, we see (from prime factorization) that we can produce infinitely many values of $n$, and hence infinitely many tuples $(a,b,c,d,e) = (n-2,n-1,n,n+1,n+2)$.