# Primes expressible as $\frac{4k^3}{(k+m)^3}+\frac{6k^2}{(k+m)^2}+\frac{4k}{(k+m)}+1$ for $k,m\geq1$

Does anyone know how to express two primes such that $$P=\frac{4k^3}{(k+m)^3}+\frac{6k^2}{(k+m)^2}+\frac{4k}{(k+m)}+1,$$ where all numbers are nonzero integers?

-
Here is an example p=5 k=1 m=1 –  Vassilis Parassidis Dec 17 '11 at 17:54

Here is a start:

$$P=\frac{4k^3}{(k+m)^3}+\frac{6k^2}{(k+m)^2}+\frac{4k}{(k+m)}+1$$

Means

$$p(k+m)^4= 4k^3(k+m)+6k^2(k+m)^2+4k(k+m)^3+(k+m)^4 \,.$$

Adding $k^4$ on both sides you get

$$p(k+m)^4+k^4=\left( k+(k+m) \right)^4 \,.$$

Thus

$$P(k+m)^4=(2k+m)^4-k^4=(2k+m-k)(2k+m+k)((2k+m)^2+k^2)$$

or

$$P(k+m)^3=(3k+m)(5k^2+4km+m^2) \,.$$

You can probably work from here by looking to the gcd $(k+m, 3k+m)$.

If $k,m$ are positive, the following is a simple continuation:

$$(3k+m) < 3(k+m)$$ $$5k^2+4km+m^2 < 5k^2+10km+5m^2=5(k+m)^2 \,.$$

Thus, $P < 10$.

Now, for each $p \in \{ 2,3,5,7 \}$ the equation

$$P(k+m)^3=(3k+m)(5k^2+4km+m^2) \,.$$

is a cubic equation in $\frac{m}{k}$, you are asking when it has a rational solution.... you can solve it numerically for each case.

Alternate route

Consider $P$ a parameter, and use the cubic formula to solve

$$4x^3+6x^2+4x+(1-p)=0 \,.$$

If $x= \frac{k}{m+k}$ is an irreducible solution, then $k|p-1$ and $m+k |2$.

Without loss of generality, you can assume that $m+k >0$.

Then $m+k =1$ or $m+k=2$.

If $m+k=1$ then we get

$$4k^3+6k^2+4k+1-p=0 \,;$$

with $k$ integer

while

If $m+k=2$ then we get

$$k^3+3k^2+4k+2-2p =0 \,.$$

with $k$ integer.

The cubic formula should help, but probably someone will see something smarter...

-