# Simultaneous Equations Challenge

#### anemone

##### MHB POTW Director
Staff member
Solve the following system in real numbers:

$$\displaystyle a^2+b^2=2c$$

$$\displaystyle 1+a^2=2ac$$

$$\displaystyle c^2=ab$$

#### Ackbach

##### Indicium Physicus
Staff member
Solve the following system in real numbers:

$$\displaystyle a^2+b^2=2c$$

$$\displaystyle 1+a^2=2ac$$

$$\displaystyle c^2=ab$$
I find it fascinating that the first equation requires $c\ge 0$. Then the second equation in conjunction with the first requires $a>0$, and incidentally tightens the $c$ inequality to $c>0$. Then the third equation, in conjunction with $c>0$ and $a>0$ requires $b>0$. Dominos!

#### Jester

##### Well-known member
MHB Math Helper
My solution
a = b = c = 1.

I have a proof that this is the only solution but I'm looking for something a little more elegant.

If we add the first two equations and then 2 times the third we end up with

$(a-b)^2+(a-c)^2+(c-1)^2 = 0$

then $a = b = c$ and $c = 1$ giving our result.

Last edited:

#### anemone

##### MHB POTW Director
Staff member
My solution
a = b = c = 1.

I have a proof that this is the only solution but I'm looking for something a little more elegant.

If we add the first two equations and then 2 times the third we end up with

$(a-b)^2+(a-c)^2+(c-1)^2 = 0$

then $a = b = c$ and $c = 1$ giving our result.
Thank you Jester for participating and your solution is so much neater and shorter than mine! Well done! My solution:

From $$\displaystyle a^2+b^2=2c$$ and $$\displaystyle c^2=ab$$, we have

$$\displaystyle a^2+\frac{c^4}{a^2}=2c\;\rightarrow\;\;a^4-2a^2c+c^4=0$$(*)

And from $$\displaystyle a^2+1=2ac$$ we square both sides and get

$$\displaystyle a^4+2a^2+1-4a^2c^2=0$$(**)

Now, subtracting the equation (*) from (**) gives

$$\displaystyle 2a^2+1-2a^2c-c^4=0$$

$$\displaystyle 2a^2(1-c)+(1-c^4)=0$$

$$\displaystyle 2a^2(1-c)+(1-c)(1+c+c^2+c^3)=0$$

$$\displaystyle (1-c)(2a^2+1+c+c^2+c^3)=0$$

Since a, b, and c >0, $$\displaystyle 2a^2+1+c+c^2+c^3 \ne 0$$ and thus it must be $$\displaystyle 1-c=0\;\rightarrow\;\;c=1$$.

Back substituting $$\displaystyle c=1$$ to the equation $$\displaystyle a^4-2a^2c+c^4=0$$ and gives

$$\displaystyle a^4-2a^2+1=0$$

$$\displaystyle a^2-1=0$$

$$\displaystyle a^2=\pm1$$ Since $$\displaystyle a>0$$, we conclude that $$\displaystyle a=1$$.

And from $$\displaystyle c=ab$$, we have

$$\displaystyle 1=1(b)\;\rightarrow b=1$$.