Prove a given number is an odd integer

[anemone]

Prove that the product of 5 positive integers $(p,\,q,\,r,\,s,\,t)$ satisfying the equation $pqrst=5(qrst+prst+pqst+pqrt+pqrs)$ is an odd integer.

 

[Opalg]

Prove that the product of 5 positive integers $(p,\,q,\,r,\,s,\,t)$ satisfying the equation $pqrst=5(qrst+prst+pqst+pqrt+pqrs)$ is an odd integer.

Should there be some extra condition here? If not, I think that the result is false.

Suppose that $p=q=r=20$ and $s=t=40$. Then $$\frac1p + \frac1q + \frac1r + \frac1s + \frac1t = \frac1{20} + \frac1{20} + \frac1{20} + \frac1{40} + \frac1{40} = \frac4{20} = \frac15.$$ Multiply out the fractions to see that $pqrst=5(qrst+prst+pqst+pqrt+pqrs)$. But each of the integers $p,q,r,s,t$ is even, and so is their product.

 

[anemone]

I am so so sorry for the late response, Opalg! I can explain...I was very busy celebrating Chinese New Year with my family and relatives in the past few days. Sorry...

I checked and argh! I typed the problem wrongly, I am deeply sorry about that...the problem should read as follow:

Prove that the number of 5 positive integers $(p,\,q,\,r,\,s,\,t)$ satisfying the equation $pqrst=5(qrst+prst+pqst+pqrt+pqrs)$ is an odd integer.

 

[anemone]

Solution of other:

Spoiler

We rewrite the given equation as

$\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}+\dfrac{1}{s}+\dfrac{1}{t}=\dfrac{1}{5}$

The number of five positive integers $(p,\,q,\,r,\,s,\,t)$ which satisfy the given relation and for which $p\ne q$ is even, because for if $(p,\,q,\,r,\,s,\,t)$ is a solution, then so is $(q,\,p,\,r,\,s,\,t)$ which is distinct from $(p,\,q,\,r,\,s,\,t)$.

Similarly the number of five positive integers which satisfy the equation and for which $r\ne s$ is also even.

Hence, it sufices to count only those five positive integers $(p,\,q,\,r,\,s,\,t)$ for which $p=q$ and $r=s$. Thus the equation reduces to

$\dfrac{2}{p}+\dfrac{2}{r}+\dfrac{1}{t}=\dfrac{1}{5}$

Here again $(p,\,p,\,r,\,r,\,t)$ for which $p\ne r$ is even.

Thus, it suffices to consider the equation

$\dfrac{4}{p}+\dfrac{1}{t}=\dfrac{1}{5}$

and show that the number of pairs $(p,\,t)$ satisfying this equation is odd.

This reduces to

$pt=20t+5p$

$(p-20)(t-5)=100$

Observe that

$100=1\times 100=2\times 50=4\times 25=5\times 20=10\times 10=20\times 5=25\times 4=50\times 2=100\times 1$

Note that no factorization of 100 as product of two negative numbers yield a positive pairs $(p,\,t)$.

Hence we get these 9 solutions. This proves that the total number of five positive integers $(p,\,q,\,r,\,s,\,t)$ satifying the given equation is odd.