What Are All the Real Solutions to the Equation \(a^4+b^4+c^4+1=4abc\)?

[anemone]

Here is this week's POTW:

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Find all real solutions of the equation $a^4+b^4+c^4+1=4abc$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

No one answered last week's problem. :(

You can find the proposed solution below:

Spoiler

The given equation can be rewritten such that in the following form:

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

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

Therefore all three terms must be zero and the solutions are hence

$(a,\,b,\,c)=(-1,\,1,\,-1),\,(-1,\,-1,\,1),\,(1,\,-1,\,-1),\,(1,\,1,\,1)$.