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Let it be four odd positive integers $a,b,c,d$. Is it possible the following relationship?

$$abcd = 2abc+bcd+cda+dab$$

Operating,

$$abc(d-2)=d(bc+ca+ab)$$

$$\frac{d}{d-2}=\frac{abc}{ab+bc+ca}$$

I got stuck at this point. Clearly, $d$ and $d-2$ are relatively primes, as they are odd, but I am not sure that this is substantial enough to get any conclusion.

Any hint / comment on how to follow would be welcomed!

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    $\begingroup$ Do you take $a=b=c=d=5$ as a solution? If you are looking for a general solution, dividing both sides by $abcd$ may help. $\endgroup$
    – player3236
    Commented Oct 7, 2020 at 18:55
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    $\begingroup$ (9,9,9,3) is another. $\endgroup$
    – cosmo5
    Commented Oct 7, 2020 at 19:07

3 Answers 3

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Dividing both sides by $abcd$:

$$1 = \frac2d+\frac1a+\frac1b+\frac1c$$

According to https://math.stackexchange.com/a/301294/435724, which lists all 147 ways to decompose $1$ a sum of 5 unit fractions, we find the relevant solutions are:

118) [3, 3, 5, 9, 45]
121) [3, 3, 5, 15, 15]
127) [3, 3, 7, 7, 21]
129) [3, 3, 9, 9, 9]
139) [3, 5, 5, 5, 15]
147) [5, 5, 5, 5, 5]

and no more. Let the repeating ones be $d$ and the rest be $a,b,c$. There are a total of $6+6+4+6+6+1 = 29$ solutions. Looking at these solutions, I have no intent to solve the question purely algebraically.

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  • $\begingroup$ thanks! I observe that all the solutions have some variable $x_n$ such that $\gcd(x_n,x_i)=x_i$ for all $i$. Could it be proven that any possible solution needs to have this property? $\endgroup$
    – Juan Moreno
    Commented Nov 10, 2020 at 12:47
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The relationship is not correct, because :

We put $a=b=c=d=1$

$Abcd= 2 abc +bcd+cda+dab $ $\Leftrightarrow$ $ 1=2+1+1+1 $

So, this relationship is not correct

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The relationship is correct over the finite field $\Bbb F_2$, because then $a=b=c=d=1$, and indeed $1=2+1+1+1$ over $\Bbb F_2$, since $2=0$. Moreover, we could cancel $d$ from the equation and obtain $$ abc=ab+ac+bc. $$

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  • $\begingroup$ If $abc=ab+bc+ba$, we would have that $\frac{d}{d-2}=1$, which is impossible for positive integers, no? $\endgroup$
    – Juan Moreno
    Commented Nov 10, 2020 at 12:38

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