MHB Simultaneous Equations Challenge

[anemone]

Solve the following system in real numbers:

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

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

$$c^2=ab$$

 

[Ackbach]

Solve the following system in real numbers:

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

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

$$c^2=ab$$

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!

 

[Mathwebster]

My solution

Spoiler

a = b = c = 1.

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

Spoiler

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.

 

[anemone]

My solution

Spoiler

a = b = c = 1.

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

Spoiler

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:

Spoiler

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

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

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

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

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

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

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

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

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

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

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

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

$$a^2-1=0$$

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

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

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