A little twist to an old problem

$\large a+b+c= \dfrac {a+b}{c}$

Find all the solutions to the diophantine equation above.

#NumberTheory

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

For (a,b,c) belong to $\mathbb{Z}$ , I have a solution..
I think the solution for (a,b,c) is of the form:

(-4-t, t, 2) for some integer t . My solution:
$ac+bc+c^2-a-b=0$ $\Rightarrow (ac-a)+(bc-b)+(c^2-1)=-1$ $\Rightarrow (c-1)(a+b+c+1)=-1$ For product of two integers to be -1 they must be of the form (-1,1) or vice versa....
In our case c-1=-1 can be easily rejected since for that value RHS is turning to undefined.. Hence
c-1=1 and a+b+c+1=-1 and hence the ordered triplet for (a,b,c) is (-4-t,t ,2) for some integer t.

Rishabh Jain - 5 years, 4 months ago

You may use the fact that $c$ divides $a+b$ .

Nihar Mahajan - 5 years, 4 months ago

Is that helpful in finding the solution?

Because I seem to have tried it before and it didn't work out for me.

Mehul Arora - 5 years, 4 months ago

Let $a+b=ck$ for some integer $k$ . Then $ck+c=k \Rightarrow c=\dfrac{k}{k+1}$ .

Nihar Mahajan - 5 years, 4 months ago

@Nihar Mahajan – Haven't you missed a=-4-t, b=t, c=2 (for some integer t) ... Hence the equation would have infinite integral solutions

Rishabh Jain - 5 years, 4 months ago

@Rishabh Jain – I just realized that $k$ can be negative , so my solution is wrong.

Nihar Mahajan - 5 years, 4 months ago

@Nihar Mahajan – Yeah... You had to take k=-2 for c to become 2..... Anyways Is my solution complete? ( I'm just a beginner in solving Diophantine equation :-})..Might be I have missed some cases too!!

Rishabh Jain - 5 years, 4 months ago

@Rishabh Jain – I have found solution when $k$ is negative. Let $k=-m$ for some positive integer $m$ , then $c=\dfrac{m}{m-1}$ . Again let $m=n(m-1)$ for some integer $n$ , so we have $\dfrac{1}{m}+\dfrac{1}{n}=1$ and the only integer solution for this is $m=n=2$ . Substituting it we have $c=2$ , $a+b=-4$ so the solution set is $(a,b,c)=(-4-t,t,2)$ for some integer $t$ .

PS your solution is shorter , because you are cool :P

Nihar Mahajan - 5 years, 4 months ago

(You should have declared if a,b,c are integers or just real) If they are real then a=b=c=2/3 is a solution

Tudor Darius Cardas - 5 years, 4 months ago

Oh yeah. That is a possible solution.

What would it be for integers?

Mehul Arora - 5 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$