# three distinct positive integers $a, b, c$ such that the sum of any two is divisible by the third

I need to determine three distinct positive integers $a, b, c$ such that the sum of any two is divisible by the third.

I tried like with out loss of generality let $a<b<c$

As, $a\mid (b+c)$ so $b+c=ak_1$ for some $k_1\in\mathbb{N}$ similarly $$a+b=k_2 c$$ $$a+c=k_3b$$ so adding them I get $k_1+k_2+k_3=2$, could anyone help me to proceed?

• I tried $1,2$ and $3$ and it seems to work. – muzzlator Jul 3 '13 at 14:42
• @Stefan : $2(a+b+c) = (a+b) + (b+c) + (c+a) = k_2c + k_1a + k_3 b$. I think that he's wrong in saying he got that equation, but it still points out that he's lost and asks for help. – Patrick Da Silva Jul 3 '13 at 14:45
• How did you get $k_1+k_2+k_3=2$? – Stefan Hamcke Jul 3 '13 at 14:45
• The mistake was in him assuming that $k_1 a + k_2 c + k_3 b = (k_1 + k_2 + k_3)(a + b + c)$ – muzzlator Jul 3 '13 at 14:46
• Also note that $a,b < c$ implies $a+b=c$. – Stefan Hamcke Jul 3 '13 at 14:51

$$ak_1-b-c=0 \ \ \ \ \ (1)$$

$$a-bk_2+c=0 \ \ \ \ (2)$$

$$a+b-ck_3=0 \ \ \ \ (3)$$

Using this, for non-trivial solution,

$$\det\begin{pmatrix} k_1 & -1 & -1 \\ 1 & -k_2 &1 \\ 1&1&-k_3\end{pmatrix}=0$$

$$\implies k_1(k_2k_3-1)-1(k_3+1)+1(1-k_2)=0$$

$$\implies k_1+k_2+k_3=k_1\cdot k_2\cdot k_3$$

From the answer by BrianM.Scott of this, we can take $k_1=1,k_2=2,k_3=3$ for any base $b$ where $b$ is a natural number $>1$

Solving $(1),(2),(3)$ we get $b=2c,a=3c\implies (a,b,c)=(3c,2c,c)$

• @CalvinLin, please observe the answer by Brian M. Scott before downvoting – lab bhattacharjee Jul 3 '13 at 16:44
• Interesting idea, though a lot of work is hidden in the second link (which arguably complicates the issue at hand). Since Brian's answer is fine, I'd undo my down vote if you edit it (since my vote is now locked). – Calvin Lin Jul 3 '13 at 16:48
• @CalvinLin, I expect thorough observation from a man of your calibre before any action (like down-voting etc.). But, I don't want to edit just for vote unless there is any mistake identified. Of course, you can request the moderator to undo the down-vote if you want – lab bhattacharjee Jul 3 '13 at 16:52
• I read your solution, and referenced it to explain my down vote. You linked to a question which specifically said digits, and that assumption was crucial in your solution. I hope you are not asking me to generalize all possible variants of a problem and consider what happens when we remove all restrictions. If the question (or your solution) didn't have the assumption of digits, I wouldn't have made the remark. As you pointed out, Brian's solution worked in the more general setting. – Calvin Lin Jul 3 '13 at 16:54
• @CalvinLin, reading other solutions of the linked problem, especially when one solution is restrictive – lab bhattacharjee Jul 3 '13 at 16:56

Since $0+0 < a + b < c + c = 2c$ and $c | a + b$, hence $a + b = c$.

Since $a < b$, hence $2a < a+b =c < 2 b$.

Since $0+ b < a + c < b + 2b$, and $b|a+c$ hence $a+c = 2b$.

Solving the system of equations, we get that $b=2a, c=3a$. It's easy to check that $\{a, 2a, 3a\}$ satisfies the conditions.

• Woah! Almost the same.:) – Inceptio Jul 3 '13 at 14:58
• @Inceptio Yours takes more work to figure out what is happening. Mine is based on the same principles of why $a+b=c$. – Calvin Lin Jul 3 '13 at 15:00
• Yes, your answer is more elegant. (+1) – Inceptio Jul 3 '13 at 15:05
• my brain is not accepting why if $a+b<2c$ and $c\mid a+b$ together imply $a+b=c$? which property/ fact of divisibility we are using here? – Marso Jul 3 '13 at 15:08
• @TaxiDriver $a+b$ is a multiple of $c$ that is between 0 and $2c$. What are all the multiples of $c$ between $0c$ and $2c$? Hint: There is only one of them. This same logic is used for the conclusion of $b < a+c < 2b$. – Calvin Lin Jul 3 '13 at 15:10

You should get $2(a+b+c)=k_1a+k_2b+k_3c$ - $(*)$

Now if $k_1=k_2=k_3=2$ we get $a=b=c$.

Suppose $k_1\le k_2\le k_3$, then we must have $k_3 \gt 2$ and $k_1 \lt 2$ else the two sides of $(*)$ cannot be equal.

Since $k_1$ is a non-zero positive integer less than $2$ is must be $1$.

So we have $a=b+c$, and $b+(b+c)=k_2c$, and $c+(b+c)=k_3b$ so that $$b=\frac {k_2-1}2c, \text{ } c=\frac {k_3-1}2b$$ whence $$(k_2-1)(k_3-1)=4$$ and we have $k_2=k_3=3$ with $b=c$, which is not allowed, or $k_2=2, \text { } k_3=5$.

In this case $c=2b$ and $a=b+c=3b$ gives the family of solutions suggeted in the comments.

• I've left some details to be checked. – Mark Bennet Jul 3 '13 at 14:59

First, note that $0<a<b<c$ and $c|a+b$ implies $a+b=c$.

Since $b|a+c=2a+b$ we have $\beta b=2a+b$ and similarly $\alpha a=a+2b$ where $\alpha,\beta\in\mathbb N$.

Now $a<b$ implies $\beta b<3b$, so $\beta$ must equal $2$ and therefore $2a=b$.

It follows that $c=a+b=3a$.

So the solutions are $(a,2a,3a)$ for any $a\in\mathbb N_+$

we are given $a<b<c$, $0<a+b<c+c=2c$ and $c\mid a+b$ so $a+b=c$, now $2a+b=a+c$ and we know $b\mid a+c$ so $b\mid 2a$. Since $2a<2b$ so $b=2a$ and hence $c=a+b=3a$ so $n,2n, 3n$ will work for any $+ve$ integer $n$