find a,b,c

Albert

Well-known member
a,b,c are positive integers
(a-b)(b-c)(c+a)=- 90
(a-b)(b+c)(c-a)=42
(a+b)(b-c)(c-a)=- 60
find a,b,c

anemone

MHB POTW Director
Staff member
a,b,c are positive integers
(a-b)(b-c)(c+a)=- 90
(a-b)(b+c)(c-a)=42
(a+b)(b-c)(c-a)=- 60
find a,b,c
My first solution:
Consider $(a-b)(b+c)(c-a)=42=2\cdot7\cdot3$, we know we must have $c>a>b>0$ and by applying AM-GM inequality to the terms $(a-b),\,(b+c),\,(c-a)$ gives $\dfrac{(a-b)+(b+c)+(c-a)}{3}\ge \sqrt[3]{42}$ or simply $c \ge 5.21$.

An educated guess by observing again the equality of $(a-b)(b+c)(c-a)=42$ reveals that $c=6,\,b=1$ and hence $a=3$ work!

Second solution:
If we are to add the last two given equations, we have

$(c-a)((a-b)(b+c)+(a+b)(b-c))=42-60$

Expand and simplifying we get

$b(c-a)^2=9=1\cdot3^2$

This tells us $b=1$, $c-a=3$ and replacing $c-a=3$, $b=1$ into $(a-b)(b+c)(c-a)=42$ we have

$(a-1)(1+c)(3)=42$

$(a-1)(1+3+a)=14$

$a=-6$ or $a=3$

Since $a>0$, $\therefore a=3$ and hence $c=6$.

Note that if we take $b=9$ and $c-a=1$, we ended up with $a=11,\,b=9,\,c=12$ but this set of solution doesn't satisfy the condition of the given equation.

Last edited:

Well-known member
a,b,c are positive integers
(a-b)(b-c)(c+a)=- 90
(a-b)(b+c)(c-a)=42
(a+b)(b-c)(c-a)=- 60
find a,b,c

Staff member

anemone

MHB POTW Director
Staff member

http://mathhelpboards.com/challenge-questions-puzzles-28/find-b-c-5637.html

Given all the problem generously posted by some of our members, this is bound to happen from time to time.
Oh My...I was totally unaware of that challenge problem been posted before at our site, and even if I did see it before, I didn't remember it at all.

Fortunately my solutions don't match with either kaliprasad's or Opalg's method, or else I may risk myself for being viewed as the "copycat" at our site.

Well-known member
My first solution:
Consider $(a-b)(b+c)(c-a)=42=2\cdot7\cdot3$, we know we must have $c>a>b>0$ and by applying AM-GM inequality to the terms $(a-b),\,(b+c),\,(c-a)$ gives $\dfrac{(a-b)+(b+c)+(c-a)}{3}\ge \sqrt[3]{42}$ or simply $c \ge 5.21$.

An educated guess by observing again the equality of $(a-b)(b+c)(c-a)=42$ reveals that $c=6,\,b=1$ and hence $a=3$ work!

Second solution:
If we are to add the last two given equations, we have

$(c-a)((a-b)(b+c)+(a+b)(b-c))=42-60$

Expand and simplifying we get

$b(c-a)^2=9=1\cdot3^2$

This tells us $b=1$, $c-a=3$ and replacing $c-a=3$, $b=1$ into $(a-b)(b+c)(c-a)=42$ we have

$(a-1)(1+c)(3)=42$

$(a-1)(1+3+a)=14$

$a=-6$ or $a=3$

Since $a>0$, $\therefore a=3$ and hence $c=6$.

Note that if we take $b=9$ and $c-a=1$, we ended up with $a=11,\,b=9,\,c=12$ but this set of solution doesn't satisfy the condition of the given equation.
we know we must have c>a>b>0

I do not know the basis for the same.