Hint $\rm\: mod\ 2,\:$ where $\rm\:x^2 \equiv x,\:$ the equations imply $\rm\:\color{#C00}{c\!+\!1} \equiv a\!+\!1 + b\!+\!1\equiv a+b\equiv \color{#c00}c\ $ $\Rightarrow\Leftarrow$
Alternatively, $ $ if modular arithmetic is unfamiliar, then you can instead show that
$\rm\quad a^2\! +\! b^2\:$ and $\rm\:(a\!+\!1)^2\!+(b\!+\!1)^2\:$ have equal parity, but $\rm\:c^2$ and $\rm\,(c\!+\!1)^2$ have opposite parity
To do that, notice that squaring preserves parity, since $\rm\,odd^2\! = odd,\:$ $\rm\:even^2\! = even,\,$ hence the parity of $\rm\:(a\!+\!1)^2\! + (b\!+\!1)^2$ equals that of $\rm\:a\!+\!1 + b\!+\!1 = a+b+2\:$ equals that of $\rm\:a\!+\!b,\:$ since adding $2$ preserves parity. Similarly the parity of $\rm\:(c\!+\!1)^2\:$ equals that of $\rm\:c\!+\!1\:$ differs from $\rm\:c.\:$ Therefore we can't have $\rm\:(c\!+\!1)^2 = (a\!+\!1)^2\!+(b\!+\!1)^2\:$ since the LHS has parity same as $\rm\:c\!+\!1\:$ but, as shown, the RHS has parity same as $\rm\:a^2+b^2 = c^2\:$ same as $\rm\:c.$
Remark $\ $ Notice how the above proof is shortened to a single line in my first proof, by arithmetizing it using modular arithmetic. This powerful and efficient modular approach has widespread application in number theory (and algebra), so one should learn it as soon as possible.