MHB Prove a given number is an odd integer

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The discussion focuses on proving that the number of positive integer solutions to the equation \( pqrst = 5(qrst + prst + pqst + pqrt + pqrs) \) is an odd integer. An example was provided where \( p, q, r \) are set to 20 and \( s, t \) to 40, leading to a product that is even, suggesting the need for additional conditions. The original problem was miscommunicated, and the correct assertion is about the count of solutions being odd. Clarifications and apologies were made regarding the initial confusion in the problem statement. The conversation emphasizes the importance of accurately framing mathematical problems for proper analysis.
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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.
 
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anemone said:
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.
 
I am so so sorry for the late response, Opalg! I can explain:o...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:mad:, 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.
 
Solution of other:

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.
 
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