How do I continue from here?

Hint: substitute $a=\tan(u), b=\tan(v), c=\tan(w)$, where $u+v+w=\pi$

@tree3 I'd be interested to know how that might help here.

Ellipti urve interetion. orry for the miing letter

WizardMath wrote: Ellipti urve interetion. orry for the miing letter Exactly, just counting the number of rational points on the curve.

Ankoganit wrote: Show that there are infinitely many rational triples $(a, b, c)$ such that $$a + b + c = abc = 6.$$ As usual, the problem is not original. This is Problem 11827 of the American Mathematical Monthly Volume 122 Issue No. 3. Moreover isn't this problem way too high for postals? (I don't know of any elementary solution till now) If required I can post a solution which uses elliptic curves.

yes Quote: If required I can post a solution which uses elliptic curves.

Wait a minute. I think I just found an elementary solution! Is there any mistake in my following solution? It is easy to show that if a triangle $ABC$ has rational side lengths and a rational area, the values of $\tan\tfrac{A}{2}$, $\tan\tfrac{B}{2}$, $\tan\tfrac{C}{2}$ are all rational. And given this, we also know that $\tan(2\theta)=\tfrac{2\tan\theta}{1-\tan^2\theta}$. Thus $\tan A, \tan B, \tan C$ are all rational. Rest follows!

YESMAths wrote: Wait a minute. I think I just found an elementary solution! Is there any mistake in my following solution? It is easy to show that if a triangle $ABC$ has rational side lengths and a rational area, the values of $\tan\tfrac{A}{2}$, $\tan\tfrac{B}{2}$, $\tan\tfrac{C}{2}$ are all rational. And given this, we also know that $\tan(2\theta)=\tfrac{2\tan\theta}{1-\tan^2\theta}$. Thus $\tan A, \tan B, \tan C$ are all rational. Rest follows! Nice

Computational proof: Take any two rational numbers $k$ and $l$. Then $a=\frac{2k}{k^2-1}$, $b=\frac{2l}{l^2-1}$ and $c=\frac{2(k+l)(kl-1)}{(k+l)^2-(kl-1)^2}$ are rational numbers satisfying $a+b+c=abc$. (* obviously, avoid $k=\pm 1,l=\pm 1$ and $(k,l)$ for which $(k+l)^2=(kl-1)^2$, for $a,b,c$ to be "defined" )

Ghoshadi wrote: Computational proof: Take any two rational numbers $k$ and $l$. Then $a=\frac{2k}{k^2-1}$, $b=\frac{2l}{l^2-1}$ and $c=\frac{2(k+l)(kl-1)}{(k+l)^2-(kl-1)^2}$ are rational numbers satisfying $a+b+c=abc$. (* obviously, avoid $k=\pm 1,l=\pm 1$ and $(k,l)$ for which $(k+l)^2=(kl-1)^2$, for $a,b,c$ to be "defined" ) just a liitle doubt $abc=6$ is not satisfied right?

As galav mentions,YESMaths's and Ghoshadi's approaches don't ensure $abc=6$. Also, I find it unlikely that an elementary solution exists; it can be easily seen that proving this theorem automatically implies problem D 16 in R K Guy's Unsolved Problems in Number Theory. (BTW, it's not unsolved as of now; a solution can be found here). Edit: Well, I'm an idiot. The problem that solves D 16 has the added condition that $a,b,c>0$ (which makes it significantly harder).

Here is the solution using elliptic curves.

Here is the outline of my submission that i can recollect. This should be an elementary solution(if it is correct ) We have if $a_i,b_i,c_i$ satisfy $abc=a+b+c=6$ then so do $$a_{i+1}=\dfrac{a_i(b_i-c_i)^2}{(c_i-a_i)(a_i-b_i)}, b_{i+1}=\dfrac{b_i(c_i-a_i)^2}{(a_i-b_i)(b_i-c_i)}, c_{i+1}=\dfrac{c_i(a_i-b_i)^2}{(b_i-c_i)(c_i-a_i)}$$(motivation:tangent chord method) Let $(a_0,b_0,c_0)=(1,2,3)$. Then we can show by induction that $v_2(b_n)=2n+1$. Thus, all triples are distinct and we can generate infinitely many.

@ above , your solution is awesome ( I think it should be hopefully correct ) ! EDIT :I think you typoed b1 and c1 ......

kapilpavase wrote: Here is the outline of my submission that i can recollect. This should be an elementary solution(if it is correct ) We have if $a_i,b_i,c_i$ satisfy $abc=a+b+c=6$ then so do $$a_{i+1}=\dfrac{a_i(b_i-c_i)^2}{(c_i-a_i)(a_i-b_i)}, b_{i+1}=\dfrac{b_i(c_i-a_i)^2}{(a_i-b_i)(b_i-c_i)}, c_{i+1}=\dfrac{c_i(a_i-b_i)^2}{(b_i-c_i)(c_i-a_i)}$$(motivation:tangent chord method) Let $(a_0,b_0,c_0)=(1,2,3)$. Then we can show by induction that $v_2(b_n)=2n+1$. Thus, all triples are distinct and we can generate infinitely many. Exactly the same solution here too!

@kapilpavase can you explain the solution better and it's motivation please.

kapilpavase wrote: (motivation:tangent chord method) Could you please explain the tangent chord method?

Draw a tangent to the curve and let it intersect the curve again at another point. It is easy to see from Viete's relations that the coordinates of the constructed point are rational if the previous point had rational coordinates.

@above, your solution didn't work for me when I was doing this problem as a part of the postals. And whatever you do, you will be faced with a cubic, so good luck for that. And isn't coordinate bash an elementary method btw? This is called something non elementary, though someone may disagree with me and call this elementary too!

WizardMath wrote: @above, your solution didn't work for me when I was doing this problem as a part of the postals. And whatever you do, you will be faced with a cubic, so good luck for that. And isn't coordinate bash an elementary method btw? This is called something non elementary, though someone may disagree with me and call this elementary too! That's non-elementary. Because, as mentioned in my post above - it uses elliptic curves.

Here's another way to generate the same answer as kapilpavase. Let $(a,b,c)$ be a solution and choose $(u,v)$ such that $au+v=b, bu+v=c$. Then consider solutions of the form $(x,xu+v,y)$. We have that $P(x)=x(xu+v)(6-u(x+1)-v)-6=0$. We know that $P(a)=P(b)=0$, and that the product of all three roots is $-6/u(u+1)=-abc/u(u+1)$. So the third root is $-c/u(u+1)$. Since $u=(b-c)/(a-b)$, we get that the third root is in fact $c(a-b)^2/(b-c)(a-c)$. Once we have the first element of the new triplet, the other two are easy to calculate (and easier to guess as the cyclic equivalents).

https://ocw.mit.edu/courses/mathematics/18-781-theory-of-numbers-spring-2012/lecture-notes/ refer to 26 to get an idea of this approach further.