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There are exactly four positive integers $n$ such that $\frac{(n + 1)^2}{n + 23}$ is an integer. Compute the largest such $n$.

Apr 20, 2018

#1
+985
+4

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.

Apr 20, 2018
#2
+101322
+1

Very nice, GYanggg  !!!!!

CPhill  Apr 20, 2018
#3
+985
+2

Thanks Chris!

It's always nice when someone complements your work!

GYanggg  Apr 20, 2018
#4
+101322
+2

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

Apr 20, 2018