# When $\frac{a^2}{b+c}＋\frac{b^2}{a+c}＋\frac{c^2}{a+b}$ is integer and $a,b,c$ are coprime natural numbers, is there a solution except (183,77,13)?

Given $a,b,c\in \Bbb{N}$ such that $\{a,b,c\}$ are coprime natural numbers and $a,b,c>1$. When $$\frac{a^2}{b+c}＋\frac{b^2}{c+a}＋\frac{c^2}{a+b}\in\mathbb Z\,?$$ I know the solution $\{183,77,13\}$. Is there any other solution?

• sorry, $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}$ is correct – MAEA2 Mar 11 '17 at 3:12
• sorry,it's first time I use math over flow. Thanks for approvemenrt! – MAEA2 Mar 11 '17 at 3:25
• you can edit your own questions if you make a mistake - look just right and down of the vote arrows – JMP Mar 11 '17 at 3:27
• I guess you don't regard $0$ as a natural number? Otherwise let one of $a, b, c$ be $0$ and the other two be $1$. – KConrad Mar 11 '17 at 3:27
• On the trivial side, $(3,1,1)$ is a solution. – T. Amdeberhan Mar 11 '17 at 4:39

Yes, there is another solution. The next one I found is a bit big, namely $$a = 15349474555424019, b = 35633837601183731, c = 105699057106239769.$$
This solution also satisfies the property that $$\frac{a^{2}}{b+c} + \frac{b^{2}}{a+c} + \frac{c^{2}}{a+b} = \frac{31}{21} (a+b+c),$$ which was true of $a = 13$, $b = 77$ and $c = 183$. If you clear denominators in the equation above, you find that both sides are multiples of $a+b+c$, and dividing out gives a plane cubic.
One wishes to search for rational points on this plane cubic, which by scaling can be assumed to be relatively prime integers. We also need points where $a, b, c > 0$ and $a+b+c$ is a multiple of $21$. This plane cubic is isomorphic to the elliptic curve $$E : y^2 + xy + y = x^3 - 8507979x + 9343104706,$$ and $E(\mathbb{Q})$ has rank $3$ and the torsion subgroup is $\mathbb{Z}/6\mathbb{Z}$. Let $E(\mathbb{Q}) = \langle T, P_{1}, P_{2}, P_{3} \rangle$ where $6T = 0$. I searched for points of the form $c_{1}T + c_{2}P_{1} + c_{3}P_{2} + c_{4}P_{3}$ where $0 \leq c_{1} \leq 5$ and $|c_{2}|, |c_{3}|, |c_{4}| \leq 5$. Of these $7986$ points, there are $1464$ where $a$, $b$ and $c$ are all positive, and these yield $11$ solutions (where $a < b < c$), the smallest of which is $(13,77,183)$. The second smallest is the one I gave above. It should be straightforward to prove that one can get infinitely many solutions in this way.
EDIT: I searched for solutions that satisfy $$\frac{a^{2}}{b+c} + \frac{b^{2}}{a+c} + \frac{c^{2}}{a+b} = t(a+b+c)$$ for all rational numbers $t = d/e$ with $d, e \leq 32$. This yields a number of smaller solutions. In particular, for $t = 9/5$ the rank of the curve is $2$ and one finds $$a = 248, b = 2755, c = 7227.$$ For $t = 21/17$ the rank is $2$ and one finds $$a = 35947, b = 196987, c = 401897.$$ Interestingly, $t = 31/21$ is the only case where the rank is $\geq 3$.
• I used Magma to get the isomorphism and compute generators. Sage probably would do both parts work as well (especially since $E$ has a rational $2$-torsion point). – Jeremy Rouse Mar 11 '17 at 13:36