No such positive reals exists. Suppose that the equations are satisfied simultaneously. Then we should have $x=2(c-a-b)$ so \[ a^2+b^2 = c^2 >(a+b)^2 \]which is impossible.

Suppose there exists. $a^2+ b^2 = c^2\wedge (a + x)^2+ (b +x)^2 = (c + x)^2\implies 2ax+2bx+x^2=2cx\iff a+b+\frac{x}{2}=c\iff \frac{x}{2}=c-a-b\implies c>a+b\iff c^2>a^2+b^2+2ab\iff 0>ab$ contradiction with $a,b>0$

WolfusA wrote: Suppose there exists. $a^2+ b^2 = c^2\wedge (a + x)^2+ (b +x)^2 = (c + x)^2\implies 2ax+2bx+x^2=2cx\iff a+b+x=c\iff x=c-a-b\implies c>a+b\iff c^2>a^2+b^2+2ab\iff 0>ab$ contradiction with $a,b>0$ i think you make mistake here $a+b+x=c$ this is wrong

raghoodah1m wrote: WolfusA wrote: Suppose there exists. $a^2+ b^2 = c^2\wedge (a + x)^2+ (b +x)^2 = (c + x)^2\implies 2ax+2bx+x^2=2cx\iff a+b+x=c\iff x=c-a-b\implies c>a+b\iff c^2>a^2+b^2+2ab\iff 0>ab$ contradiction with $a,b>0$ i think you make mistake here $a+b+x=c$ this is wrong it really doesnt matter because then $\frac{x}{2}=c-a-b$

Use some manipulations $\frac{1}{4}x^2 +(a+b)x+2ab=0$ $x=-2(a+b)+-2\sqrt{a^2+b^2-6ab}$

arshiya381 wrote: raghoodah1m wrote: WolfusA wrote: Suppose there exists. $a^2+ b^2 = c^2\wedge (a + x)^2+ (b +x)^2 = (c + x)^2\implies 2ax+2bx+x^2=2cx\iff a+b+x=c\iff x=c-a-b\implies c>a+b\iff c^2>a^2+b^2+2ab\iff 0>ab$ contradiction with $a,b>0$ i think you make mistake here $a+b+x=c$ this is wrong it really doesnt matter because then $\frac{x}{2}=c-a-b$ ok, thank you

crezk wrote: Use some manipulations $\frac{1}{4}x^2 +(a+b)x+2ab=0$ $x=-2(a+b)+-2\sqrt{a^2+b^2-6ab}$ now it is clear

We have, $\bigg(\frac{a}{c}\bigg)^2+\bigg(\frac{b}{c}\bigg)^2=1$ and also $\bigg(\frac{a+x}{c+x}\bigg)^2+\bigg(\frac{b+x}{c+x}\bigg)^2=1.$ But $\bigg(\frac{a+x}{c+x}\bigg)>\bigg(\frac{a}{c}\bigg)$ since $c>a,$ so $\bigg(\frac{a+x}{c+x}\bigg)^2>\bigg(\frac{a}{c}\bigg)^2$ and similarly $\bigg(\frac{b+x}{c+x}\bigg)^2>\bigg(\frac{b}{c}\bigg)^2.$ Adding, we have $\bigg(\frac{a+x}{c+x}\bigg)^2+\bigg(\frac{b+x}{c+x}\bigg)^2>\bigg(\frac{a}{c}\bigg)^2+\bigg(\frac{b}{c}\bigg)^2=1,$ a contradiction.

$\color{blue} \boxed{\textbf{SOLUTION}}$ Let, there axis positive $x$ s.t. $(a+x)^{2}+(b+x)^2=(c+x)^2$ $\implies a^2+2ax+x^2+b^2+2bx+x^2=c^2+2cx+x^2$ $\implies(2a+2b)x+x^2-2cx=0$ $\implies(2a+2b-2c)x +x^2=0$ $\implies (2a+2b-2c+x)x=0$ $\implies x=2[c-(a+b)], x=0$ but from $\textbf{Triangle Inequality,}$ $c < (a+b) \implies x \in 0,$ Negative Real Numbers So, there exists no positive real $x$ satisfying the condition $\blacksquare$