DanDumitrescu wrote: Determine all triples $(a,b,c)$ of integers that simultaneously satisfy the following relations: \begin{align*} a^2 + a = b + c, \\ b^2 + b = a + c, \\ c^2 + c = a + b. \end{align*} Add all of the equations to get $a^2+b^2+c^2=a+b+c$. Since $a^2 \ge a$ for all integers $a$, $a,b,c=0,1$. Therefore, the only solutions are $(0,0,0)$ and $(1,1,1)$.

Adding all of the equations, we have $a^2+b^2+c^2=a+b+c$. For all integers, $x^2\geq{x}$ since the y coordinate of a parabola is always greater or equal to the x coordinate. Equality holds at a=b=c=1, or a=b=c=0.

huashiliao2020 wrote: Adding all of the equations, we have $a^2+b^2+c^2=a+b+c$. For all integers, $x^2\geq{x}$ since the y coordinate of a parabola is always greater or equal to the x coordinate. Equality holds at a=b=c=1, or a=b=c=0. you just copied me

ReaperGod wrote: huashiliao2020 wrote: Adding all of the equations, we have $a^2+b^2+c^2=a+b+c$. For all integers, $x^2\geq{x}$ since the y coordinate of a parabola is always greater or equal to the x coordinate. Equality holds at a=b=c=1, or a=b=c=0. you just copied me Well sorry I didn't look at your solution before typing mine, and there are slight differences but this is a very common approach

Different approach, but still a very basic one. $a^2+2a, b^2+2b, c^2+2c$ is all equal to $a+b+c$. Since a quadratic equation cannot have 3 distinct roots, there exists not necessarily different $m,n\in\mathbb{Z}$ such that two of $a,b,c$ is $m$ and the other is $n$. Obviously $m^2+2m=n^2+2n$, thus $m=n$ or $m+n=-2$. If $m=n$, $m^2+2m=3m$ and $m=0,1$. Thus $(a,b,c)=(0,0,0),(1,1,1)$. If $m\neq n$, $m^2+2m=m+m+n=m-2$ and this has no integer roots. The only possible $(a,b,c)$ is $(0,0,0),(1,1,1)$.

What about this? Subtract the first two to get $(a-b)(a+b-2)=0$ so $a=b$ or $a+b=2$. Similarly for the other two. I mean, there are only $8$ cases, so yea