Egyptian Fractions & Square

First of all, the equation can be written as $\frac{ab + bc + ca}{abc} = \frac{x^2}{n}$ . I will now prove that $\gcd(abc,ab + bc + ca) = 1$ :

Suppose that $d | abc , ab + bc + ca$ . Then $d | abc + bc^2 + c^2 a \implies d | bc^2 + c^2a \implies d | b^2c^2 + bc^2 a \implies d | b^2 c^2$ . If $d$ does not divide $bc$ then $d$ is divided by a higher power of a prime that divides $bc$ but then $d$ would not divide $abc$ . This is a contradiction so $d | bc$ . By symmetry (or by repeating the procedure) we also get that $d | ab, bc , ca$ . So $d | \gcd(ab,bc) = b$ and $d | \gcd(bc, ac) = c$ so $d | \gcd(b,c) = 1$ so $d=1$ .

Because the expression of a positive rational number as a quotient of relatively prime positive integers is unique, $\frac{ab + bc + ca}{abc} = \frac{x^2}{n} \implies ab + bc + ca = x^2, abc = n$ . Notice that this guarantees that $\gcd(n, x^2) = 1$ . So any solution of the equation $ab + bc + ca = x^2$ with relatively prime variables will yield a solution to the original problem by simply taking $n = abc$ . Solving this diophantine equation in general terms is relatively easy as $ab + bc + ca = x^2 \implies (b+a)(c+a) = x^2 + a^2$ so by computing any particular value of $x^2 + a^2$ and then factorizing it one would find a solution to this equation, but this will not guarantee that the solution will meet our conditions.

The rest of my solution is heuristic, there is probably a better solution :

As $x^2 \equiv 0,1 \mod 4$ we can break it down into cases. Suppose $ab + bc + ca \equiv 0 \mod 4$ . Suppose that $a$ is even, then $b,c$ are odd. But we have $a(b + c) + bc \equiv 0 \mod 4 \implies bc \equiv 0 \mod 4$ so at least another variable has to be even, this is a contradiction. Suppose no variable is even, then we would have $\pm 1 \pm 1 \pm 1 \equiv 0 \mod 4$ . Regardless of how you choose your signs, this is impossible. This guarantees that $x^2 \equiv 1 \mod 4$ .

Suppose that none of the variables $a,b,c$ are even. If $a,b,c \equiv -1 \mod 4$ then $ab + bc + ca \equiv 3 \mod 4$ which is impossible so suppose $c=1$ , then $ab + b + a \equiv 1 \mod 4$ . If $a \equiv -1$ then $-1 \equiv 1 \mod 4$ . Impossible so suppose $a \equiv 1$ , then $2b \equiv 0 \mod 4$ which implies $b$ is even, a contradiction. This means that at least one (and at most one) of the variables has to be even. Let's suppose that $a$ is even. Then $a(b+c) + bc \equiv 1 \mod 4 \implies bc \equiv 1 \mod 4 \implies b \equiv c \mod 4$ . In short, all solutions must have one even variable and the two other variables must be both congruent to either $1$ or $-1 \mod 4$ .

Using the previous result to guide the search, if we restrict ourselves to small numbers (specifically, numbers no larger than 10) then by trial and error one finds the solution $a=10, b=3, c=7$ with $x = 11$ by making a list of relatively prime triples among the naturals $2,3,4,...,10$ that also satisfy the previous congruences. These are:

$(2,3,7), (2,5,9), (4,3,7), (4,5,9), (8,3,7), (8,5,9), (10,3,7)$ . You can easily find them by noticing that $a = 2,4,6,8,10$ and $b,c$ are either $3,7$ or $5,9$ . With this computation we find that the only solution in small variables is the one mentioned with $a+b+c = 10 + 3 + 7 = 20$ . We know that this is the smallest value of $a+b+c$ because any other solution must have at least two variables with two digits so $a + b + c > 10 + 10 = 20$ or have a variable bigger than $10$ so $a + b + c > 10 + 3 + 7 = 20$ .