Here is a different solution: First, if $p\mid c$ then simply take $a=0$. Hence suppose $p\nmid c$. Note that by Euler's criterion, $a^{\frac{p-1}{2}} = (\frac{a}{p})$, and thus, $a^{\frac{p+1}{2}}\equiv a\pmod{p}$ if $a$ is a quadratic residue, and is equal to $-a$ if $a$ is a quadratic non-residue. Now, suppose first $c$ is a quadratic residue modulo $p$. Then, taking $a=0$ suffices. Hence assume $c$ is not a quadratic residue modulo $p$. If $p\equiv 1,5\pmod{8}$, then $-1$ is a quadratic residue modulo $p$. In this case, take $a=-c$. Then, $-c$ is not a quadratic residue modulo $p$, and thus, $(-c)^{\frac{p+1}{2}}\equiv c\pmod{p}$ from the paragraph above. Now assume $p\equiv -1\pmod{4}$. In this case, $-1$ is not a quadratic residue, hence, $-c$ is a quadratic residue. It now suffices to show there is a $k$ such that $k$ is a quadratic residue, whereas $k+1$ is not. To see how this helps, assume $k$ is such a number. Then, take $a=-kc-c=-(k+1)c$. $a$ is not a quadratic residue modulo $p$, hence $a^{\frac{p+1}{2}}\equiv -a\equiv kc+c\pmod{p}$, whereas $a+c =-kc$ which is a quadratic residue modulo $p$, and thus $(a+c)^{\frac{p+1}{2}}\equiv a+c\equiv -kc\pmod{p}$ concluding the case. To show such a $k$ exists, assume that it does not, that is, for each quadratic residue $k$, $k+1$ is also a quadratic residue. But then, since $1$ is a quadratic residue, we get inductively that everybody in $\mathbb{Z}_p$ is a quadratic residue, which yields a contradiction.

For $p=3$, we can manually check that for $c=0$ or $1$, we can take $a=0$, and for $c=2$, we can take $a=2$. For $p>3$, if $c$ is a quadratic residue, then take $a=0$ and, by Euler's criterion, $c^{(p+1)/2}\equiv c \pmod{p}$ and we are done. Suppose now that $c$ is not a residue. Then if we find $a$ which is a nonresidue such that $a+c$ is a residue, we are done since the left hand side is then $-a+(a+c)\equiv c \pmod{p}$. Suppose that such $a$ doesn't exist. This implies that the mapping $x \mapsto x-c$ is a bijection from the set of residues to the set of nonresidues. Let $x$ be a nonresidue. Then $xc$ is a residue, and $xc-c=(x-1)c$ is a nonresidue, which implies $x-1$ is a residue. This implies that the mapping $x \mapsto x-1$ is a bijection from the set of nonresidues to the set of residues. This means that in the sequence $1, 2, \ldots, p-1$, no two consecutive numbers are nonresidues, and no two are residues. Since $1$ is a residue, $2$ is then a nonresidue, and then $3$ is a residue, and then $4$ is not a residue, which is a contradiction.

In the solution, all the things are done in modulo $p$. By Euler's Criteria we have $x^{\frac{p+1}{2}} \equiv \left(\dfrac{x}{p}\right) \cdot x \pmod{p}$. If $c$ is a quadratic residue, then we can take $a=0$. Now assume $c$ is not a quadratic residue. If $p \equiv 1 \pmod{4}$, then $-c$ is also not a quadratic residue and so taking $a = -c$ works. Assume $p \equiv 3 \pmod{4}$. Then it is clear that $-c$ is a quadratic residue modulo $p$. Consider the numbers $-c , -2c , -3c , \dots , -(p-1)c$. Since $c \neq 0$ and there are $\frac{p-1}{2}$ non-zero quadratic residues, there exists $\frac{p+1}{2}$ of them which are not a quadratic residue. Let $j \geq 2$ be the first number such that $-jc$ is not a quadratic residue. Then plugging in $a = -jc$ and using that $-(j-1)c$ is a quadratic residue we have $jc + -(j - 1)c \equiv c \pmod{p}$, so we are done.

Basically the same idea as above - posting for storage. Assume that $p \nmid c$ as otherwise simply take $a = 0$. Notice by Euler's Criterion that we want an $a \in \mathbb{F}_p$ such that $a+c$ is a quadratic residue while $a$ is not. Assume for the sake of contradiction that there are no such $a$. Then for any quadratic residue $b \in \mathbb{F}_p$, $b-c$ is also a quadratic residue which is not possible as $gcd(p,b) = 1$ would mean that we have that all elements of $\mathbb{F}_p$ are quadratic residues because $ck$ is surjective in $\mathbb{F}_p$ where $k$ also varies in $\mathbb{F}_p$ which is a contradiction. $\blacksquare$