I will tell the fundamental things for solution.If $a^2+2b+c=x^2$ and the others are squares of $y,z$, notice that $x,y,z$ cant be $>=a+2,b+2,c+2$ same time.and if $x=a$ then $a,b,c=0$ so this case is trivial. Thus we get at least one of these integers must be equal to $a+1,b+1,c+1$ .By this way you can solve it.

Darn what a terrible problem. The answer is $(1,1,1)$ and permutations of $(127, 106, 43)$. WLOG $a \ge b,c$. Then note that we need $a^2 + 2b + c = (a+1)^2 \implies 2b + c = 2a + 1$ so $c$ is odd so let $c = 2c' + 1$. Then we have $b + c' = a$. Now we have two cases. If $b \ge c$, then note that $b > \frac{2}{3}a$ so $b^2 + 2c + a < b^2 + 3.5b < (b+2)^2$ and therefore $b^2 + 2c + a = (b+1)^2$ and therefore \[4c' + 2 + a = 2b + 1 \implies 5c' + 1 = b\] Therefore we arrive at a parametrization of $(a,b,c) = (6c' + 1, 5c' + 1, 2c' + 1)$ and its not terribly hard to verify that $c' = 0, c' = 21$ are the only solutions. If $b < c$, then $c > \frac{2}{3}a$ and thus $c^2 + 2a + b < c^2 + 4c < (c+2)^2$ so $c^2 + 2a + b = (c+1)^2$ and therefore \[2a + b = 4c' + 3 \implies 3b = 2c' + 3\] and so again we arrive at a parametrization by letting $c' = 3k$ of $(a,b,c) = (5k + 1,2k + 1,6k + 1)$. But then $c > a$, contradiction so $(1,1,1)$ and permutations of $(127, 106, 43)$ are the only solutions. EDIT : Oops I read it as positive integers. For nonnegative the same argument works unless at least $2$ of $a,b,c$ are $0$, in which case it is obvious that the third is $0$ as well which gives $(0,0,0)$ as another solution.

Wow problem is really hard but they are nonNegative so $(0,0,0)$ is solution too.Thanks for solution

Its just extension of a APMO 2011 problem , see http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2274367&sid=f2b44b4182578bc8c311bc22c67929e1#p2274367

Cute problem; almost as cute as Lol_man000 We claim the only solutions are $(1,1,1), (0,0,0), (43, 127, 106)$ and its relative permutations. WLOG, let $a\le b\le c.$ Observe then that $2a+b\le 3c < 4c+4,$ and $2a+b\ge 0,$ giving that $$c^2\le c^2+2a+b\le (c+1)^2.$$If $c^2 = c^2+2a+b,$ then $\boxed{a=b=c=0}$ is a solution. Otherwise, $2a+b=2c+1.$ Moreover, $$b^2+2c+a = b^2+3a+b-1 \leq b^2+4b-1 < b^2 + 4b +4 = (b+2)^2.$$Hence, $$b^2\leq b^2+2c+a\leq (b+1)^2.$$In the former, we have the same solution set as before, so $2c+a=2b+1.$ Combining with the previous condition gives that $b=3a-2$ $2c=5a-3$ Meaning that the first expression is $$a^2< a^2+\frac{17}{2}a-\frac{11}{2} < a^2+10+25=(a+5)^2.$$Hence, setting $a^2+\tfrac{17}{2}a-\tfrac{11}{2}$ equal to $(a+1)^2, (a+2)^2, (a+3)^2, (a+4)^2,$ it is clear the only nonnegative integer solutions for $a$ are $a=43, 1.$ These give solution sets $\boxed{(43, 127, 106)}$ and $\boxed{(1,1,1)},$ respectively. $\blacksquare$

cmsgr8er wrote: Cute problem; almost as cute as Lol_man000 We claim the only solutions are $(1,1,1), (0,0,0), (43, 127, 106)$ and its relative permutations. WLOG, let $a\le b\le c.$ Observe then that $2a+b\le 3c < 4c+4,$ and $2a+b\ge 0,$ giving that $$c^2\le c^2+2a+b\le (c+1)^2.$$If $c^2 = c^2+2a+b,$ then $\boxed{a=b=c=0}$ is a solution. Otherwise, $2a+b=2c+1.$ Moreover, $$b^2+2c+a = b^2+3a+b-1 \leq b^2+4b-1 < b^2 + 4b +4 = (b+2)^2.$$Hence, $$b^2\leq b^2+2c+a\leq (b+1)^2.$$In the former, we have the same solution set as before, so $2c+a=2b+1.$ Combining with the previous condition gives that $b=3a-2$ $2c=5a-3$ Meaning that the first expression is $$a^2< a^2+\frac{17}{2}a-\frac{11}{2} < a^2+10+25=(a+5)^2.$$Hence, setting $a^2+\tfrac{17}{2}a-\tfrac{11}{2}$ equal to $(a+1)^2, (a+2)^2, (a+3)^2, (a+4)^2,$ it is clear the only nonnegative integer solutions for $a$ are $a=43, 1.$ These give solution sets $\boxed{(43, 127, 106)}$ and $\boxed{(1,1,1)},$ respectively. $\blacksquare$ You can't assume WLOG $a \leqslant b \leqslant c$ as it's not symmetrical; it's cyclic. The best you can do is assume $c = \max(a, b, c)$ and then tackle cases $a \leqslant b$ and $a > b$ seperately.

did this a while ago but didn't post The only solutions are $(0,0,0)$, $(1,1,1)$, and $(127,106,43)$ (and of course cyclic permutations). These work. Now we prove they are the only solutions. Assume that $a,b,c$ are not all zero. If two of the variables are $0$, then we have $a, 2a$ are perfect squares, so $a = 0$. Now assume that at most one of the variables is zero. WLOG that $a = \max(a,b,c)$. We have $a^2 + 2b + c > a^2$ and $a^2 + 2b + c \le a^2 + 2a + a < (a+2)^2$, so $a^2 + 2b + c = (a+1)^2$, meaning $2b + c = 2a + 1$. Case 1: $c > b$. Then we have $c^2 + 2a + b$ is a perfect square. We have \[ c^2 + 2a + b = c^2 + (2b + c - 1) + b = c^2 + c + 3b - 1 \le c^2 + 4c - 1 < (c+2)^2 ,\]implying that $c^2 + 2a + b = (c+1)^2$, so $2a + b = 2c + 1$. Hence $3b + c - 1 = 2c + 1\implies c = 3b - 2$. However, we have $2b + c = 2a + 1$, so $5b - 3 = 2a\implies a = \frac{5b - 3}{2}$. Since $a \ge c$, we have $\frac{5b - 3}{2} > 3b - 2$, which is impossible. Case 2: $b \ge c$. We have $b^2 + 2c + a = b^2 + 2c + \frac{2b + c - 1}{2} \le b^2 + 2b + \frac{3b - 1}{2} < (b+2)^2$, so $b^2 + 2c + a = (b+1)^2$, implying $2c + a = 2b + 1$. Letting $c = 2k + 1$ yields that $a = b + k$ and $4k + 2 + b + k = 2b + 1$, so $b = 5k + 1$. Hence, we have $(a,b,c) = (6k+1, 5k+1, 2k+1)$. Now, $(2k+1)^2 + 2(6k + 1) + 5k + 1$ is a perfect square. This can be rewritten as $4k^2 + 21k + 4$ is a perfect square. Since $(2k+2)^2 \le 4k^2 + 21k + 4 < (2k+6)^2$, we have $4k^2 + 21k + 4 \in \{ (2k+2)^2, (2k+3)^2, (2k+4)^2, (2k+5)^2\} $. Thus, either $21k + 4 \in \{8k + 4, 12k + 9, 16k + 16, 20k + 25\}$. The middle two cases don't give integer solutions. The first case gives $k = 0$ and last case gives $k = 21$. If $k = 0$, then $a = b = c = 1$. If $k = 21$, then $(a,b,c) = (127,106,43)$, as desired.