-1
$\begingroup$

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?

  • 0
    Here is an example p=5 k=1 m=12011-12-17

1 Answers 1

1

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...