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It is intuitive that $a=\pm \frac{c}{2}$, with $-\frac{c}{2}\leq b\leq \frac{c}{2}$ or vice-versa are solutions to the problem. Can I get to these solutions without dividing the expression in all the cases (i.e. $a$ and $b$ being positive or negative and $|a|$ being greater than or less than $|b|$)? Nothing wrong with that, I'm just looking for other way of solving this problem. I tried using the fact that $|x|=\sqrt{x^2}$ but the expression quickly becomes very complex with the two variables.

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    $\begingroup$ "without dividing the expression in all the cases " What's wrong with dividing the expression in all the cases? $\endgroup$ – fleablood Apr 10 at 23:09
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There are several standard methods: intervals (cases), squaring, graphical.

Let's do squaring: $$(|a+b|+|a-b|)^2=c^2 \iff \\ 2a^2+2b^2+2|a^2-b^2|=c^2 \iff \\ (2|a^2-b^2|)^2=(c^2-2(a^2+b^2))^2 \iff \\ 4a^2+4b^2-8a^2b^2=c^4-4(a^2+b^2)c^2+4a^2+4b^2+8a^2b^2 \iff \\ c^4-4(a^2+b^2)c^2+16a^2b^2=0 \Rightarrow \\ c^2=2(a^2+b^2)\pm \sqrt{(2(a^2+b^2))^2-16a^2b^2}=2a^2+2b^2\pm \sqrt{4(a^2-b^2)^2} \Rightarrow \\ c^2=2a^2+2b^2\pm 2|a^2-b^2|\stackrel{WLOG: \ |a|\ge |b|}{=} 4a^2;4b^2 \Rightarrow \\ a=\pm \frac c2; |b|\le |a| \Rightarrow -\frac c2\le b\le \frac c2.$$

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What's wrong with doing all cases?

Case 1: $a+b \ge 0; a-b \ge 0$.

Then $|a+b|+|a-b| = (a+b)+ (a-b) = c\ge 0$. And $a =\frac c2$ and $b$ mayb be anything in $[-\frac c2, \frac c2]$.

Case 2: $a+b \ge 0; a-b < 0$.

Then $|a+b| + |a-b| = (a+b) + (b-a) = c \ge 0$. And $b = \frac c2$. $a -\frac c2 < 0$ and $a+\frac c2 \ge 0$ so $a$ may be anything in $[-\frac c2, \frac c2)$.

Case 3: $a+b < 0$ and $a-b \ge 0$.

Then $|a+b| + |a-b| = -a-b + a-b = -2b = c \ge 0$ so $b = -\frac c2$ and $a\ge b$ and $a < -b$ so $a$ may be anything in $[-\frac c2, \frac c2)$.

Case 4: $a+b < 0$ and $a-b < 0$.

Then $|a+b| + |a-b| = -a -b +b-a=-2a = c\ge 0$ so $a = -\frac c2$ and $b< \frac c2$ and $b > -\frac c2$ as $b$ may be anything in $(-\frac c2, \frac c2)$.

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So solutions are $a =\pm \frac c2; b\in [-\frac c2, \frac c2]$ and $b = \pm \frac c2; a\in [-\frac c2, \frac c2]$.

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WLOG, we may assume $a,b \ge 0$. Also $a \ge b$.

Therefore, $c = |a+b|+|a-b| = a+b + a-b = 2a$, and so $a=\frac c2$ and $b \le a=\frac c2$.

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