+239 There are exactly four positive integers $n$ such that \[\frac{(n + 1)^2}{n + 23}\] is an integer. Compute the largest such $n$.
+983 Hi Rollingblade!
The first step in this problem is to divide the numerator by its denominator.
We have:
\(\begin{align*} \frac{(n+1)^2}{n+23} &= \frac{n^2 +2n + 1}{n+23}\\ &= \frac{n^2 + 23n - 21n+1}{n+23}\\ &= \frac{n^2 +23n}{n+23} -\frac{21n}{n+23} + \frac{1}{n+23}\\ &=\frac{n(n+23)}{n+23} - 21\cdot \frac{n+23-23}{n+23}+ \frac{1}{n+23}\\ &=n - 21\left(\frac{n+23}{n+23} - \frac{23}{n+23}\right)+ \frac{1}{n+23}\\ &= n - 21 + \frac{21\cdot 23}{n+23}+ \frac{1}{n+23}\\ &=n-21 + \frac{484}{n+23}. \end{align*} \)
This means that n+23 must be a factor of 484 = 2^2 *11^2
The largest factor of 484 is 484.
So n = 484 - 23 = 461.
I hope this helps.
+129852 Here's one more way using polynomial division
(n + 1)^2 n^2 + 2n + 1
_______ = ___________
n + 23 n + 23
n - 21
n + 23 [ n^2 + 2n + 1 ]
n^2 + 23n
__________________
-21n + 1
-21n - 483
___________
484
n - 21 + 484
________
n + 23
And note that, as GYanggg found, the largest n that makes the last fraction an integer is when n = 461
