How to obtain three consecutive integers?

[Math100]

Homework Statement

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.

Relevant Equations

None.

Let $a, a+1$ and $a+2$ be the three consecutive integers.
Then
\begin{align*}
&5^{2}\mid a\implies a\equiv 0\pmod {25}\\
&3^{3}\mid (a+1)\implies a+1\equiv 0\pmod {27}\implies a\equiv 26\pmod {9}\\
&2^{4}\mid (a+2)\implies a+2\equiv 0\pmod {16}\implies a\equiv 14\pmod {16}.\\
\end{align*}
Applying the Chinese Remainder Theorem produces:
$n=25\cdot 27\cdot 16=10800$.
This means $N_{1}=\frac{10800}{25}=432, N_{2}=\frac{10800}{27}=400$ and $N_{3}=\frac{10800}{16}=675$.
Now we have $432x_{1}\equiv 1\pmod {25}, 400x_{2}\equiv 1\pmod {27}$ and $675x_{3}\equiv 1\pmod {16}$.
Observe that
\begin{align*}
&432x_{1}\equiv 1\pmod {25}\implies 7x_{1}\equiv 1\pmod {25}\\
&\implies 49x_{1}\equiv 7\pmod {25}\implies -x_{1}\equiv 7\pmod {25}\\
&\implies x_{1}\equiv 18\pmod {25},\\
&400x_{2}\equiv 1\pmod {27}\implies -5x_{2}\equiv 1\pmod {27}\\
&\implies -25x_{2}\equiv 5\pmod {27}\implies 2x_{2}\equiv 5\pmod {27}\\
&\implies 28x_{2}\equiv 70\pmod {27}\implies x_{2}\equiv 16\pmod {27},\\
&675x_{3}\equiv 1\pmod {16}\implies 3x_{3}\equiv 1\pmod {16}\\
&\implies 15x_{3}\equiv 5\pmod {16}\implies -x_{3}\equiv 5\pmod {16}\\
&\implies x_{3}\equiv 11\pmod {16}.\\
\end{align*}
Since $x_{1}=18, x_{2}=16$ and $x_{3}=11$,
it follows that $x\equiv (0+26\cdot 400\cdot 16+14\cdot 675\cdot 11)\pmod {10800}\equiv 270350\pmod {10800}\equiv 350\pmod {10800}$.
Thus, $a=350, a+1=351$ and $a+2=352$.
Therefore, the three consecutive integers are $350, 351$ and $352$.

 

[fresh_42]

The answer is correct, but how did you come to choose $(5,3,2)$? And there is a typo in the second line where you wrote $\pmod 9$ instead of $\pmod{27}.$

 

[Math100]

The answer is correct, but how did you come to choose $(5,3,2)$? And there is a typo in the second line where you wrote $\pmod 9$ instead of $\pmod{27}.$

I chose $(5, 3, 2)$ from the book's answer, since it says $5^{2}\mid 350, 3^{3}\mid 351$ and $2^{4}\mid 352$. But do you know why it should be $(5, 3, 2)$? Because honestly, I do not know why. It seems like because all $2, 3$ and $5$ are prime numbers.

 

[fresh_42]

I chose $(5, 3, 2)$ from the book's answer, since it says $5^{2}\mid 350, 3^{3}\mid 351$ and $2^{4}\mid 352$. But do you know why it should be $(5, 3, 2)$? Because honestly, I do not know why. It seems like because all $2, 3$ and $5$ are prime numbers.

We are asked to find $r^2\,|\,a\, , \,s^3\,|\,(a+1)\,|\,t^4\,|\,(a+2).$

I wonder why they all have to be pairwise distinct. I assume there is a reason, but I haven't looked into it. If so, then $r,s,t$ shouldn't be too far apart to obtain a solution that can be computed without computers. It makes sense to choose $t$ as small as possible in order to keep $a$ small. That gives us $t^4=16\,|\,(a+2).$ Then $s=3$ is the next small number to keep $s^3$ and $a$ possibly small. The only question left is then, should we choose $r=4$ or $r=5?$ But from $r=4$ we would get $16|a$ and $16|(a+2)$ which is impossible.

So $(5,3,2)$ is the smallest possible triple with distinct numbers. And it worked. There are probably more solutions. It is just the smallest one.