If two of the three numbers $a, b, c$ are equal, we can assume without loss of generality (oBdA) due to cyclic symmetry that $a=b$. This transforms the given left equation to $a^{2}+p b=a^{2}+p c$, from which $b=c$ follows directly because $p \neq 0$, thus satisfying the claim. Henceforth, we can assume $a \neq b \neq c \neq a$. Rearranging the equations yields $p=\frac{b^{2}-a^{2}}{b-c}=\frac{c^{2}-b^{2}}{c-a}=\frac{a^{2}-c^{2}}{a-b}$. Multiplying the three terms and simplifying gives $p^{3}=-(a+b)(b+c)(c+a)(1)$.
Among the three given numbers, at least two are even or at least two are odd by the pigeonhole principle; their sum is therefore divisible by 2. Hence, the product on the right side of (1) is even. It follows that $p=2$ and therefore $(a+b)(b+c)(c+a)=-8$. (2) If two of the parentheses in (2) are equal, we can assume without loss of generality (oBdA) that $a+b=b+c$. It follows directly that $a=c$ and hence the equality of all three given numbers.
If exactly one of the parentheses is odd (equal to $\pm 1$), then $(a+b)+(b+c)+(c+a)$ is on the one hand odd, and on the other hand equal to $2(a+b+c)$, which is even - a contradiction!
Therefore, the cases ( $\pm ; \mp 1 ; 8$ ) remain to be examined (up to cyclic permutation).
From $a+b=1, b+c=-1, c+a=8$ it follows that $a=5, b=-4, c=3$, which contradicts the first given equation with $25-8 \neq 16+6$.
From $a+b=-1, b+c=1, c+a=8$ it follows that $a=3, b=-4, c=5$, which contradicts the first given equation with $9-8 \neq 16+10$.
Therefore, only $a=b=c$ is possible. Indeed, $a^{2}+p a=a^{2}+p a$ holds.
Hints: From $a>b$ it does not follow that $a^{2}>b^{2}$ and vice versa, because $a$ or $b$ can be negative.
The given equations are not symmetric in $a, b$, and $c$, but only cyclic. Therefore, it is not permissible to assume without loss of generality (oBdA) that $a>b>c$.
The claim can be generalized to all positive natural exponents $n$.