Parag's integers

Expanding the top of the fraction gives $$\frac {n^2+2n+1}{n+7}$$ Dividing using either long division or synthetic division gives $$n-5+\frac {36}{n+7}$$ $n-5$ will always be an integer so we have to force $\frac {36}{n+7}$ to be an integer. There are 9 total factors of 36 so there are 9 n's for which $n+7$ will be one of those factors, but we can also make $n+7$ equal to the negative factors of 36, so there are 18 total values of $n$ for which $\frac {36}{n+7}$ will be an integer.

We develop the first step Term $(n+1)^{2}$ $\Rightarrow$ $n^{2}+2n+1$

Second Step We Make a Synthetic Division, So We have:

$n-5$ + $\frac{36} {n+7}$ , Now we must find the $36$ For Dividers $n +7$ Divide $36$ , $A$ Does the dividers $36$ $\Rightarrow$ $A$ = $1$ , $2$ , $3$ , $4$ , $6$ , $9$ , $12$ , $18$ , $36$ , Since $9$ Dividers Positive and $n$ can Have More dividers in $9$ negative , Total $18$ Whole Numbers !

Using polynomial long division, it can be seen that: $\frac{(n+1)^2}{n+7} = n-5+\frac{36}{n+7}$ Therefore, for $\frac{(n+1)^2}{n+7}$ to be an integer, we can see that $\frac{36}{n+7}$ must also be an integer. This is achieved when $n+7$ is a factor of $36$ . The factors of $36$ are $1, 2, 3, 4, 6, 9, 12, 18, 36$ , so $n + 7$ can be any one of these values, or the negative equivalent. Hence, the total number of possible values for $n$ is: $9 + 9 = \fbox{18}$

$n+7 | n^2 + 2n + 1$

$n+7 | n^2 + 7n + 1 - 5n = n(n+6) = 1 -5n$

$n+7| 1-5n$

$n+7 | 1-5n + 5(n+7)$

$n+7 | 1+ 35 = 36$ . Therefore n+7 is a factor of 36. We note that $36 = 3^2 * 2^2$ Hence, $36$ has a total of $(2+1)(2+1) = 3 * 3 = 9$ positive divisors and thus 18 integral divisor. Also for a distinct divisor $d$ of 36, we get a distinct $d-7$ and hence a distinct value of n. Therefore, the total number of integers for which $n+7 | n^2 + 2n + 1=$ number of divisors of 36 $= \boxed{18}$

Note that $\begin{aligned} (n + 1)^2 &\equiv n^2 + 2n + 1 \pmod{n + 7} \\ &\equiv n^2 + 2n + 1 - n(n + 7) \pmod{n + 7} \\&\equiv 1 - 5n \pmod{n + 7} \\ &\equiv 1 - 5n + 5(n + 7) \pmod{n + 7} \\ &\equiv 36 \pmod{n + 7} \end{aligned}$ Hence $n + 7 \mid 36$ , now if $n + 7 \in \mathbb{Z}$ , then $n \in \mathbb{Z}$ , hence it suffices to count the number of divisors of $36$ , note that $36 = 2^2 \cdot 3^2$ , hence there are $2 \cdot 3 \cdot 3 = \boxed{18}$ such integers by the combinatorial rule of product.

$N=\frac{(n+1)^2}{n+7}$ may be written as $n-5+\frac{36}{n+7}$ . $N$ is an integer whenever both $n-5$ and $\frac{36}{n+7}$ are integers. For all integers $n$ , $n-5)$ is always an integer. Thus, we wish to find all values of $n$ such that $\frac{36}{n+7}$ is also an integer.

Clearly $(n+7)|36$ . But, $36=2^2 \cdot 3^2$ has $(2+1)\cdot(2+1)=9$ positive factors. Because we do not confine ourselves with only the set of positive integers as factors, then we can also find $9$ negative factors giving a total of $18$ . For the record, the values of $n$ desired are: $-6, -5, -4, -1, 2, 5, 11, 29, -8, -9, -10, -11, -13, -16, -19, -25,$ and $-43$ .

Just see that you have to write (n+1)^2 in the form of (n+7)(n+a)+b where a and b are integers and then [(n+1)^2/(n+7)]=(n+a)+b/(n+7)....Now only check how many positive or negative integer [taking the value of (n+7)] can divide the number "b" ...The answer is the number of positive and negative divisors of the integer "b".Here b=36...So we get 1,-1,2,-2,3,-3,4,-4,6,-6,9,-9,12,-12,18,-18,36,-36(total 18)...

(n+1)^2/n+7=(n+7-6)^2/n+7=(n-5)+(36/n+7), no of possible values of n=no of both positive and negative factors of 36 =18.

We use polynomial long division to write $\frac{(n+1)^2}{n+7}$ as $(n-5) + \frac{36}{n+7}$ this quantity is an integer if and only if $\frac{36}{n+7}$ is integral which it is whenever $n+7$ is a not necessarily positive divisor of 36. By the standard formula $36=2^2*3^2$ has $(2+1)(2+1)= 9$ positive factors, and so it has 18 total integral factors. Each factor yields a unique value of $n$ , so the answer is $\boxed{18}$

$(n+1)^2 = n^2 + 2n + 1 = n^2 + 2n -35 +36 = (n+7)(n+5) + 36$ $\frac{(n+1)^2}{n+7} = \frac{(n+7)(n+5) + 36}{n+7} = n+5 + \frac{36}{n+7}$

Hence $(n+7)|36$ There are a total of 18 integer factors of 36 (note: negative factors are counted) and every factor leads to a corresponding n.

So there are 18 integers n.

$\frac{(n+1)^{2}}{n+7}$ = $\frac{ n^{2}+2n+1}{n+7}$ = $\frac{(n+7)(n-5)+36)}{n+7} = n-5+\frac{36}{n+7}$ Now for the given expression to be a integer $n+7$ must be a factor of $36$ .

$36=2^{2} * 3^{2}$ So there are a total of 18 factors of 36( 9 positive and 9 negative).Hence we have 18 different values of n+7 or 18 different values of n

The expression is $\frac{n^2+2n+1}{n+7}$ . This is just polynomial division with an excluded value of $n=-7$ . The result is $n-5+\frac{36}{n+7}$ . Since the set of integers is closed under addition and subtraction, there is now an equation of sorts to work with. $integer-integer+\frac{36}{n+7}=integer$ . Reducing more shows that $n-5$ could me eliminated for the purpose at hand, so $\frac{36}{n+7}$ is set to equal an integer. This means that $36$ divided by $n+7$ is "even", so $n+7$ is a factor of $36$ . $36$ has 9 factors (determined through counting or by formula), so it seems that there are $9$ different quantities for $n+7$ . However, this only assumed positive values. Seeing that $n+7$ can be negative means that the answer is multiplied by $2$ (since each factor now has a negative friend). Checking that $n\neq-7$ (confirmed), the final answer is $18$ .

The key with this method is to find the remainder of the fraction (which turns out to be 36). You can of course use long division, OR you can force the numerator to have terms similar to the denominator (n+7):

$(n+1)$ ^2 = $([n+7]-6)$ ^2 = $(n+7)$ ^2 - $12 \times (n+7)$ + 36

When you divide by the denominator, (n+7):

$\frac {(n+7)^2 -12 \times (n+7) + 36}{(n+7)^2}$ =(n+7) - 12 + $\frac {36}{(n+7)^2}$

The first terms are always integers for all values of n. In order for the final fraction to be an integer, the magnitude of the denominator must be less than or equal to the numerator. Otherwise it's value is a fraction of 1.

$|n + 7| < 36$

$-36 < (n+7) < 36$

Since 36 = ${2^2} \times {3^2}$ , |n+7| must equal ${2^a} \times {3^b}$ in order to obtain integer results. Where both “a” and “b” must be 0, 1, or 2. Therefore, there are $3 \times 3$ =9 combinations.

However, each factor can be negative for: \(2 \times 9) = 18 integer values. Finally, for each unique integer value of (n+7), there is a unique integer value for "n". Therefore, the result is also:

\(\boxed {18 \text{ integers}}\)

Manipulate the numerator as 1. (n+1)^{2} = (n+7)^{2} - 12 (n+4). Hence the problem is reduced to 2. The problem is now reduced to finding n for which 12 (n+4)/(n+7) is an integer ..

A simple MATLAB code does the computation for you, giving you 18 as the answer :) count=0; for i = -100:100 A = 12*(i+4); B = i+7; if rem(A,B) == 0 i count = count + 1; end end count