High School Integer n: Solving the Equation ##\dfrac{n^2+3}{2n+4}## for Integers

[anemone]

Find all integers n such that $\dfrac{n^2+3}{2n+4}$ is an integer as well.

 

[anuttarasammyak]

Spoiler

It is obvious that n is odd integer, so say n=2m+1
The given formula becomes
m-\frac{1}{2}-\frac{\frac{7}{4}}{m+\frac{3}{2}}
So the RHS last term is half integer so
p(m)=\frac{\frac{7}{2}}{m+\frac{3}{2}}is an odd integer.
|m+\frac{3}{2}|\leq\frac{7}{2}
-5\leq m \leq 2
Only p(-5)=-1,p(-2)=-7,p(-1)=7,p(2)=1 are odd integers which result
n=-9,-3,-1,5

 

[Opalg]

If n^2 + 3 = k(2n+4) then n^2 - 2kn + (3-4k) = 0. The discriminant of that quadratic must be a perfect square, say k^2 + 4k-3 = m^2. Then (k+2)^2 - 7 = m^2. The only squares that differ by 7 are 9 and 16, so k+2 = \pm4. Thus either k = 2 and n^2 - 4n - 5 = 0, giving n = -1 \text{ or }5; or k= -6 and n^2 + 12n + 27 = 0, giving n = -3 \text{ or }-9.

 

[bob012345]

Graphically, the relation looks like the following graph if one extends it to the continuous regime. Interestingly, the integer n values are equally spaced on either side of the pole at x=−2 as −2±1 and −2±7.

 

[bob012345]

I also noticed that the lines $y=x±3$ intersect the curve $\dfrac{x^2+3}{2x+4}$ exactly at the integer solution points -9,-3,-1 and 5.

If we let $\dfrac{n^2+3}{2n+4} = n±3$ we get the four integer solutions quickly because the solutions are the intercepts of the lines with the original function. This can be generalized as;

$$\dfrac{n^2+2k+1}{2n+4} = n±(k+2) \quad k=0,1,2,3...$$
The solutions are $$n=\{-2k-7, -3,-1, 2k+3\}$$ and the original case is recovered with $k=1$.

EDIT: for very large values of $k$ this method does not get all the cases.

 

[anuttarasammyak]

Graphically, the relation looks like the following graph if one extends it to the continuous regime. Interestingly, the integer n values are equally spaced on either side of the pole at x=−2 as −2±1 and −2±7.

y=\frac{x^2+3}{2x+4}=\frac{1}{2}(x+2+\frac{7}{x+2})-2
So we see the graph is untisymmetric to the line $y=-2$ and $x=-2$

 

[anuttarasammyak]

(cont.)
say z=x+2
z+\frac{7}{z}=2n even number.
z^2-2nz+7=0 should be
(z-1)(z-7)=0 or
(z+1)(z+7)=0

 

[bob012345]

I leave it so someone else to solve the even cases; $$\dfrac{n^2+2k }{2n+4} \quad k= 1,2,3...$$

 

[anuttarasammyak]

I leave it so someone else to solve the even cases;

As for the extended problem : Find all integers n such that $\frac{n^2+d}{2n+4}$ with integer d, is an integer as well, we should get quadratic equation similar to #7
z^2-2mz+d+4=0
If d+4 is prime number we would get answers similarly.
If d+4 is square number we would get z is $\pm$ square root of it.
More in general if we can factorize
d+4=MN
where M and N are both odd or even, that provides the answer.
As for your interest of even d, it is any multiple of 4 so that both M and N are even.

 

[bob012345]

As for the extended problem : Find all integers n such that $\frac{n^2+d}{2n+4}$ with integer d, is an integer as well, we should get quadratic equation similar to #7
z^2-2mz+d+4=0
If d+4 is prime number we would get answers similarly.
If d+4 is square number we would get z is $\pm$ square root of it.
More in general if we can factorize
d+4=MN
where M and N are both odd or even, that provides the answer.
As for your interest of even d, it is any multiple of 4 so that both M and N are even.

Please clarify your process in #7. Where does the $2n$ come from?

$$z+\frac{7}{z}=2n$$

Also, for $d+4=MN$, if d is even, don't both M and N have to be even?

 

[anuttarasammyak]

Please clarify your process in #7. Where does the 2n come from?

In the formula of #6 due to the coefficient $\frac{1}{2}$, $z+\frac{7}{z}$ is an even number so that y is an integer.

Also, for d+4=MN, if d is even, don't both M and N have to be even?

M,N are solutions of the quadratic equation of z. M+N =2m is even, i.e. they are both odd or even.

 

[bob012345]

I agree the even cases only work for multiples of 4;

$$f(n)=\dfrac{n^2+4k }{2n+4} \quad k= 1,2,3...$$ where;

$$n=\{-2(k+2), -4,0, 2k\}$$ and the values of the relation are;

$$ f(n)=\{-4-k,k\}$$

EDIT: Again, for large values of $k$ this does not get all possible cases. See post #13 by @anuttarasammyak.

 

[anuttarasammyak]

I think there are more cases than you show
e.g.
k+1=\frac{M}{2}\frac{N}{2}
for k=119 (M/2,N/2)={(1,120),(2,60),(3,40),(4,30),(5,24),(6,20),(8,15),(10,12)} for M<N

 

[bob012345]

I think there are more cases than you show
e.g.
k+1=\frac{M}{2}\frac{N}{2}
for k=119 (M/2,N/2)={(1,120),(2,60),(3,40),(4,30),(5,24),(6,20),(8,15),(10,12)} for M<N

Well, I did miss some! According to Wolfram Alpha, there are 32 integer solutions for k=119.

https://www.wolframalpha.com/input?i=+++(x^2++476)/(2x+4)+=+m+

I'm not sure what you mean by

$$k+1=\frac{M}{2}\frac{N}{2}$$ or how it works to generate solutions to the original problem. For example, for k=119, what is (1,120)? How is it a solution to the original problem?

Did you meant solutions to this?

$$ \dfrac{n^2+4*119 }{2n+4} $$

 

[anuttarasammyak]

Well, I did miss some! According to Wolfram Alpha, there are 32 integer solutions for k=119.

https://www.wolframalpha.com/input?i=+++(x^2++476)/(2x+4)+=+m+

Did you meant solutions to this?

Yes. 32 solutions correspond with 16 solutions in post #13 doubled by putting minus sign.

For example, for k=119, what is (1,120)?

z=x+2=2, 240
x=0, 238
They are in your wolfram solution list. Explicitly
n=\pm 2q -2where q is a devisor of k+1
k+1 \equiv 0(mod\ q)