Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
Here is an example p=5 k=1 m=1 –  Vassilis Parassidis Dec 17 '11 at 17:54

1 Answer 1

up vote 1 down vote accepted

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...

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.