What 4 values for n in $\frac{(n+1)^2}{n+23}$ gives integer values

Question

Question What 4 values for n in $\frac{(n+1)^2}{n+23}$ gives integer values

reminder

Understand the fact that the question calls for an integer number n

My step

Well for a starter i factored out the numerator to get $n^2+2n+1$ and i have no clue what to do from there

Answer 1

Note that $$\begin{align} & \frac{(n+1)^2}{n+23}\\ & =\frac{n^2+2n+1}{n+23}\\ & =\frac{n^2+23n+1-21n}{n+23}\\\ & =n+\frac{1-21n}{n+23}\\ & =n+\frac{-21(n+23)+484}{n+23}\\ &=n-21+\frac{484}{n+23}\end{align}$$

So the values of $n$ suitable are $461,-507,-1,-12,-21,-67,21,98$ and ??.

I hope you will be able to fill it.

Answer 2

Hint

Use that

$$\dfrac{n^2+2n+1}{n+23}=n-21+\dfrac{484}{n+23}.$$

Note that $n-21$ is integer. So, you need that $$\dfrac{484}{n+23}$$ to be an integer.

Answer 3

just a Hint.

$$\frac{(n+1)^2}{n+23}=\frac{n(n+23)-21n+1}{n+23}$$

$$=n+\frac{-21(n+23)+484}{n+23}$$

$$n-21+\frac{484}{n+23}$$

with $484=4*121=4*11*11$

Answer 4

$$\dfrac{(n+1)^2}{n+23}=\dfrac{(n+23-22)^2}{n+23}=n+23-2\cdot44+\dfrac{22^2}{n+23}$$

Now the divisors of $22^2=2^2\cdot11^2$ are $\pm2^a11^b$ where $0\le a\le2,0\le b\le2$