Number Theory Homework: Find 3 consecutive integers...

Question

I have this problem assigned for homework, and I'm a bit confused as to how to solve it:

Obtain three consecutive integers, the first of which is divisible by a square, the second by a cube, and the third by a fourth power (other than $1^2, 1^3, 1^4$).

I've started like this, but haven't been able to get very far:

$x\equiv 0\pmod{a^2}$,

$x+1\equiv 0\pmod{b^3}$,

$x+2\equiv 0\pmod{c^4}$, some $a,b,c\in\mathbb{N}$.

Thus, we have:

$x\equiv 0\pmod{a^2}$,

$x\equiv -1\pmod{b^3}$,

$x\equiv -2\pmod{c^4}$.

I tried using the Chinese Remainder Theorem at this point but I was having difficulty considering everything is in terms of $a,b,c$...

The book's answer is $5^2\mid 350, 3^3\mid 351, 2^4\mid 352$.

Thanks!

Answer 1

HINT : Try to find an integral value say Y such that $2^2 | Y$, $3^3 | (Y + 1)$ , $4^4|(Y + 2)$ Then obtain 3 congruence relations and apply "Chinese Remainder Theorem".

Answer 2

With a few reasonable assumptions, an answer can be found with a minimum of brute force work. Note first that for each square, cube, and fourth power to be found, if said number divides one of the consecutive numbers, then each prime factor of that number also divides that particular number. Note also that three consecutive numbers, which I will represent as $y-1,y,y+1$, will have one member divisible by $3$, and one or two that are even.

I will make the simplifying assumption that among the numbers that are solutions, one of them will have $3$ as a prime factor, and another will have $2$ as a prime factor. This is not necessarily the only possible path to solutions, but it seems a reasonable path to solutions of modest magnitude. In this case, the product of the three consecutive numbers, $y^3-y$, will be divisible by $72$, since one of the numbers will have a factor of $3^k$ where $k$ is $2,3,4$, and there will have to be at least $3$ factors of $2$ among the numbers (at least $2^3$ if $2\mid y$ and at least $2^3$ if $2\mid (y+1)(y-1)$). Finally, there will be at least one more prime $p$ such that $p^2\mid (y^3-y)$. I will make the further simplifying assumption that $p^2\mid (y-1)$ so as to avoid the necessity to look at $p^3\mid y$ or $p^4\mid (y+1)$ for large values of $p$.

Using a simple spreadsheet, I looked at $\frac{y^3-y}{72p^2}$ for $1<y\le 500$ and $p=5,7,11,13$ to identify candidates for $y$. Choosing only results that were integers, I further culled the results for those where $p^2\mid (y-1)$.

For $p=13$ no candidates were found. For $p=11$, candidates $243,485$ were identified, but ruled out by examination with regard to the divisibility requirements. For $p=7$, candidates $99,197$ were identified, but ruled out by examination with regard to the divisibility requirements. For $p=5$, candidates $251,351,451$ were identified. Of these, only $351$ met the divisibility requirements.

So within the constraints of my assumptions, up to $500$, the only solution is $(y-1),y,(y+1)=350,351,352$.