Can Positive Integers Satisfy the Equation $ab+bc+ca=1+5\sqrt{a^2+b^2+c^2}$?

[anemone]

Solve for positive integers the equation $ab+bc+ca=1+5\sqrt{a^2+b^2+c^2}$.

 

[Opalg]

Solve for positive integers the equation $ab+bc+ca=1+5\sqrt{a^2+b^2+c^2}$.

[sp]Let $u = a+b+c$ and $v = bc + ca + ab$. Then $u^2 = (a+b+c)^2 = a^2 + b^2 + c^2 + 2(bc + ca + ab) = a^2 + b^2 + c^2 + 2v.$ Therefore $a^2 + b^2 + c^2 = u^2 - 2v$.

So the given equation tells us that $v-1 = 5\sqrt{u^2 - 2v}$. Thus $(v-1)^2 = 25(u^2 - 2v)$, which gives $v^2 + 48v + 1 = 25u^2.$

Complete the square to get $(v+24)^2 - 575 = (5u)^2,$ or $(v+24)^2 - (5u)^2 = 575.$ Factorise this as the difference of two squares: $(v+24 +5u)(v+24 - 5u) = 575.$

So $v+24 +5u = p$ and $v+24 -5u = q$, where $p$ and $q$ are integers whose product is $575 = 5^2\cdot23.$ Subtract the second equation from the first to get $10u = p-q.$ Thus $p-q$ must be a multiple of $10$. The only way that can happen with $pq = 5^2\cdot23$ is if $p = 5\cdot23$ and $q=5$. Thus $p-q = 10\cdot11$ and so $u = 11$. The corresponding value of $v$ is $v=36$.

Therefore $a+b+c = 11$, and $a^2+b^2+c^2= u^2 - 2v = 49$. That can only happen if one of $a,b,c$ is odd and the other two are even. This cuts down the possible values for the triple $\{a,b,c\}$ to a very few possible cases to check. The only one for which $a^2+b^2+c^2 = 49$ is $\{3,2,6\}$, and that does indeed provide a solution to the problem.

Conclusion: the numbers $a,b,c$ are $2,3,6$ (in any order).[/sp]

 

[anemone]

Thanks Opalg for your neat solution!

Here is another quite similar approach with Opalg that is proposed by other:

Spoiler

Clearly $ab+bc+ca>6$, so if we rewrite the given equation as follows:

$ab+bc+ca-1=5\sqrt{a^2+b^2+c^2}$

$(ab+bc+ca-1)^2=25(a^2+b^2+c^2)$

And we let $ab+bc+ca=5k+1$ where $k>0$ and $k$ is an integer so that $a^2+b^2+c^2=k^2$, we then get:

$\begin{align*}(a+b+c)^2&=a^2+b^2+c^2+2(ab+bc+ca)\\&=k^2+2(5k+1)\\&=(k+5)^2-23\end{align*}$

Therefore

$(k+5)^2-(a+b+c)^2=23$

$(k+5-a-b-c)(k+5+a+b+c)=1(23)$

Since $a,\,b,\,c,\,k$ are positive integers, it must be

$k+5-a-b-c=1$ and $k+5+a+b+c=23$

Summing up the two equations gives $k=7$.

So $a+b+c=11$ and $ab+bc+ca=36$.

Suppose WLOG, that $c>b>a$, clearly $a\le 3$. We there are three cases to consider:

$a=1$ gives $b+c=10$, $bc=26$ and this has no solution.

$a=2$ gives $b+c=9$, $bc=18$ so $b=3,\,c=6$.

$a=3$ gives $b+c=8$, $bc=12$ and this has no solution.

We can conclude by now that the solutions are hence

$(a,\,b,\,c)=(2,\,3,\,6),\,(2,\,6,\,3),\,(3,\,2,\,6),\,(3,\,6,\,2),\,(6,\,2,\,3),\,(6,\,3,\,2)$