I wonder how to solve this system of equations:
\begin{cases}a = \left(b-c\right)^{2}\\b = \left(a-c\right)^{2}\\c = \left(a-b\right)^{2}\end{cases}
Asked
Active
Viewed 67 times
0
-
1Welcome to MSE! What attempts have you done so far? – Jessie Mar 16 '21 at 10:39
-
I am assuming that you mean besides the trivial one of $(0,0,0)$. – user2661923 Mar 16 '21 at 10:43
-
The easy ones are $(0,0,0),(1,1,0),(1,0,1),(0,1,1)$ – Henry Mar 16 '21 at 10:47
-
Are $a,b,c$ required to be integers? – user2661923 Mar 16 '21 at 10:52
2 Answers
1
We have $$a - b = (b-c)^2 - (a-c)^2 = (b -c -a+c)(b-c+a - c) = (b-a)(b+a - 2c)$$ Assuming $a \not = b$, $$1 = -(b+a - 2c) = -b-a+2c \implies c = \frac{1+a+b}{2}$$Plugging in the last equation, $$\frac{1+a+b}{2} = (a-b)^2$$ Note that $a,b\ge 0$. Solving this quadratic equation when $a \ge -\frac{9}{16}$, $$b = \frac{4a - \sqrt{16a + 9} + 1}{4}, b = \frac{4a + \sqrt{16a + 9} + 1}{4}$$ Now use these solutions and the first equation to find $a = 0,b=1$ or $a = 1,b = 0$. If $a = b$ then $c = 0$ and $$a = b^2 \\ b = a^2 $$ which implies $a = b = 1$ or $a = b = 0$.
S.H.W
- 4,379
-
If $a=b=2$ and $c=0$ then $a\not = (b-c)^2$. Instead if $a=b$ then $c=0$ and $a=b^2=a^2$ so $a=0$ or $1$. – Henry Mar 16 '21 at 11:01
-
-
You are an old user, so I want to just remind you that it is not the ideal practice to post answers to without-context questions. – ultralegend5385 Mar 16 '21 at 11:09
-
@ultralegend5385 You're right. I tried to leave most of calculations(which is solving the quadratic equation) to OP. – S.H.W Mar 16 '21 at 11:14
-
You have also not fully solved the $a\not=b$ case: there are only two possible values of $(a,b)$ in that case – Henry Mar 16 '21 at 11:18
-
-
Thank you for help. However, I am still not sure how you get "b". Could you please explain it a little more in depth. – user10203585 Mar 16 '21 at 19:37
-
@user10203585 Expand $\frac{1+a+b}{2} = (a-b)^2$ and use quadratic formula to solve for $b$. – S.H.W Mar 16 '21 at 23:36
0
Hint for a possible approach:
- Substitute $a=(b-c)^2$ into $b=(a-c)^2$
- Expand this into an quartic equation involving $b$ and $c$
- That can be factorised into two quadratics involving $b$ and $c$
- So you can get four possible expressions for $b$ in terms of $c$
- You can then use each of these in $c=(a-b)^2$ to get corresponding values of $c$ (messy but simplifies nicely), and thus working backwards of $b$ and $a$
- so you end with a set of solutions for $(a,b,c)$
Henry
- 157,058