Find the sum....!

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Find the sum of all positive integers $\large{n}$ for which

$\large{\dfrac{(n+1)^2}{n+23}}$ is a positive integer.

The answer is 799.

1 solution

Patrick Corn
Jan 28, 2014

From $(n+1)^2 \equiv 0$ mod $(n+23)$ , we substitute $n \equiv -23$ to get $(-23+1)^2 \equiv 0$ mod $(n+23)$ , so $(n+23)|484$ . The factors of $484 = 2^2 \cdot 11^2$ are $1,2,4,11,22,44,121,242,484$ . Only the last four are big enough to be $n+23$ . So this gives $n = 21,98,219,461$ , which sum to $\fbox{799}$ .

Nice solution. One may avoid congruences to reach $n + 23 | 484$ in the following way.

$n+23 | n^2 + 2n + 1$

or, $n+23 | n^2 + 23n - 21n +1$

or, $n+23 | n(n+23) - 21n + 1$

or, $n+23 | 1 - 21n$

or, $n+23 | 1 - 21n + 21(n+23)$

or, $n+23 | 1 - 21n + 21n + 483 = 484$

Sagnik Saha - 7 years, 4 months ago

fantastic

sujit kumar - 7 years, 3 months ago

Find the sum....!

The answer is 799.

1 solution

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