Solve $abc+bcd+cda+dab=abcd+3$ in $\mathbb{N}$. 
Solve $abc+bcd+cda+dab=abcd+3$ in the set of positive integers.

I tried this - by rearranging we get $bcd-3=abcd-abc-abd-acd=a(bcd-bc-bd-cd)$ therefore $bcd-3$ is divisible by $a$. Similarly, we get that $cda-3$ is divisible by $b$, $dab-3$ is divisible by $c$ and that $abc-3$ is divisible by $c$. What do I do with this information? How do I go on?
 A: First of all, note (as I mentioned in a comment) that if more than one of $a,b,c,d$ is even, all of the terms on the LHS will be even (since each one omits only one term), but the RHS will be odd (being $3$ plus an even number), so they can't be equal.
Next, divide both sides by $abcd$: this yields $\dfrac1a+\dfrac1b+\dfrac1c+\dfrac1d=1+\dfrac3{abcd}$. Since the equation is clearly symmetric in its variables, we can assume WLOG that $a\leq b\leq c\leq d$. The first and biggest thing to note is that we must have $\dfrac1a+\dfrac1b+\dfrac1c+\dfrac1d\gt 1$; this will imply some bounds. For instance, $a\lt 4$, because otherwise the sum on the LHS is bounded above by $\frac14+\frac14+\frac14+\frac14=1$.  We'll get bounds on the other values on a case-by-case basis below.
First, if $a=1$ then we have the equation $\dfrac1b+\dfrac1c+\dfrac1d = \dfrac3{bcd}$, or $cd+db+bc=3$; solving in terms of $b$ yields $b=\dfrac{3-cd}{c+d}$, which immediately implies $c,d\leq 2$, and 'by inspection' the only solution in whole numbers turns out to be $b=c=d=1$.
Next, if $a=2$ then we can't have $b=2$ or $b=4$ (since there can be only one even number among $a,b,c,d$), and if $b\geq 7$ then $\dfrac1a+\dfrac1b+\dfrac1c+\dfrac1d\leq\dfrac12+\dfrac17+\dfrac17+\dfrac17\lt 1$; this implies $b=3$ or $b=5$. Consider the $b=3$ case first; then the original equation becomes $\dfrac12+\dfrac13+\dfrac1c+\dfrac1d=1+\dfrac1{2cd}$, or $\dfrac1c+\dfrac1d=\dfrac16+\dfrac1{2cd}$, or $6(c+d)=cd+3$. From the second of these we must have $c\leq 12$ (elsewise $\displaystyle\frac1c+\frac1d\leq\frac16$); from the third we get $\displaystyle d=\frac{6c-3}{c-6}$. This implies that $c=5$ doesn't work (since $d$ would be negative); $c=7$ yields $d=39$, $c=9$ yields $d=17$; and $c=11$ yields $d=\frac{63}5$ (which isn't a solution since $d$ must be integral).
In the $a=2,b=5$ case, we must have $c=5$ because $\dfrac12+\dfrac15+\dfrac17+\dfrac17\lt 1$; but now consider the original equation $abc+bcd+cda+dab=abcd+3$. With $b=c=5$ the LHS must be divisible by $5$ since each term is, but the RHS cannot be because $abcd$ is a multiple of $5$ and $3$ is not, so $abcd+3$ isn't either.
Now, if $a=3$, then we can either have $b=3$ or $b=4$; $b=5$ is ruled out since $\dfrac13+\dfrac15+\dfrac15+\dfrac15\lt1$.  $a=3,b=3$ yields the equation $\dfrac1c+\dfrac1d=\dfrac13+\dfrac1{3cd}$ or $3(c+d)=cd+1$; similarly to the above this yields $c\lt 6$ and $d=\dfrac{3c-1}{c-3}$, giving $\{c=4, d=11\}$ and $\{c=5, d=7\}$ as solutions.
Finally, we can eliminate the case $a=3,b=4$: we would need $c,d\geq 5$ so $\dfrac1a+\dfrac1b+\dfrac1c+\dfrac1d\leq \dfrac13+\dfrac14+\dfrac15+\dfrac15\lt 1$.
Putting this all together, we get the complete set of solutions as Will Jagy found in his answer: $\{1,1,1,1\}$, $\{2,3,7,39\}$, $\{2,3,9,17\}$, $\{3,3,4,11\}$, and $\{3,3,5,7\}$.
Incidentally, similar "Egyptian fraction" equations and arguments arise in classifying tilings of the regular surfaces (the plane, sphere, and hyperbolic plane); see e.g. https://johncarlosbaez.wordpress.com/2012/02/05/archimedean-tilings-and-egyptian-fractions/ for a nice discussion of this.
A: seems to be finite number of solutions, no evident way to get new solutions out of old
Mon Mar 19 10:35:26 PDT 2018
 a: 1    b: 1    c: 1    d: 1
 a: 7    b: 5    c: 3    d: 3
 a: 11    b: 4    c: 3    d: 3
 a: 17    b: 9    c: 3    d: 2
 a: 39    b: 7    c: 3    d: 2
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