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anemone
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Find all integers n such that ##\dfrac{n^2+3}{2n+4}## is an integer as well.
[tex]y=\frac{x^2+3}{2x+4}=\frac{1}{2}(x+2+\frac{7}{x+2})-2[/tex]bob012345 said: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.
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 #7bob012345 said:I leave it so someone else to solve the even cases;
Please clarify your process in #7. Where does the ##2n## come from?anuttarasammyak said: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
[tex]z^2-2mz+d+4=0[/tex]
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
[tex]d+4=MN[/tex]
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.
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.bob012345 said:Please clarify your process in #7. Where does the 2n come from?
M,N are solutions of the quadratic equation of z. M+N =2m is even, i.e. they are both odd or even.bob012345 said:Also, for d+4=MN, if d is even, don't both M and N have to be even?
Well, I did miss some! According to Wolfram Alpha, there are 32 integer solutions for k=119.anuttarasammyak said:I think there are more cases than you show
e.g.
[tex]k+1=\frac{M}{2}\frac{N}{2}[/tex]
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 said: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+
Yes. 32 solutions correspond with 16 solutions in post #13 doubled by putting minus sign.bob012345 said:Did you meant solutions to this?
z=x+2=2, 240bob012345 said:For example, for k=119, what is (1,120)?
The equation ##\dfrac{n^2+3}{2n+4}## is a rational expression that involves a variable, n, in the numerator and denominator. It is a quadratic equation because it contains a variable with an exponent of 2.
To solve this equation for integers, you will need to use algebraic techniques such as factoring, completing the square, or the quadratic formula. These methods will help you find the values of n that make the equation true.
Yes, this equation can have multiple solutions. Since it is a quadratic equation, it can have up to two solutions. These solutions may be integers or fractions, depending on the specific values of n that make the equation true.
Yes, there are some restrictions on the values of n in this equation. The denominator, 2n+4, cannot equal 0 because division by 0 is undefined. Therefore, n cannot equal -2. Additionally, since we are solving for integers, n must be a whole number.
You can check your solution to this equation by plugging it back into the original equation. If the resulting expression is true, then your solution is correct. You can also graph the equation and see if your solution falls on the graph, or use a calculator to evaluate the equation for your solution and see if it equals the original expression.