Are there three integers such the nontrivial sums are squares and three are squares of consecutives? 
Do there exist such three different positive integers $a, b, c$ that the following numbers: $a+b, b+c, c+a, a+b+c$ are all squares of integers and three of them ($a+b, b+c, c+a, a+b+c$) are squares of consecutive integers?

My first thought is to order the numbers $a,b,c$. Let $a<b<c$ and $a+b=k^2$, $b+c=l^2$, $c+a=m^2$, $a+b+c=n^2$. Then $n^2>l^2>m^2>k^2$ so $n>l>m>k$. Now we can consider two variants:
(i)  $\quad n=l+1=k+2$
(ii) $\quad l=k+1=l+2$
I don't know what to do from this place...

In the previous post, i miss understood the question.
We need to find $a,b,c$ such that $a+b,a+c,b+c,a+b+c$ all are square number and three of them are consecutive square numbers. 
Which means with $d,e$ integers $a+b,a+c,b+c,a+b+c$ are equal to $(d-1)^2,d^2,(d+1)^2 ,e^2$. since $a,b,c$ are positive so $a+b+c >a+b,a+c,b+c$ which means that $a+b+c =(d+1)^2 $ or $a+b+c = e^2$, and since the other three equations are symmetric it does not matter how to order them, so with out loss of generality let :
$a+b=(d-1)^2$ and $a+c = d^2$ and $b+c=(d+1)^2$ and $a+b+c=e^2$.
Summing togther the first three equations we get $a+b+a+c+b+c = 2(a+b+c)= 2e^2=(d-1)^2+d^2 +(d+1)^2$.
Solving the equation $2e^2 = (d-1)^2 + d^2 +(d+1)^2$ we arrive at $2e^2=3d^2+2$
Solving this equation in Wolfram for integers gives that for any positive integer number $n$ the following formulas :
$d_n =   \frac{\left(5+2 \sqrt{6}\right)^n-\left(5-2 \sqrt{6}\right)^n}{\sqrt{6}}$ and $e_n = \frac{1}{2} \left(\left(5-2 \sqrt{6}\right)^n+\left(5+2 \sqrt{6}\right)^n\right) $
Now substituting these values instead of $d,e$ in the equations $a+b=(d-1)^2,a+c=d^2,b+c=(d+1)^2,a+b+c=e^2$ gives that for any positive integer number $n$ the following formulas :
$a_n=\sqrt{\frac{2}{3}} \left(5-2 \sqrt{6}\right)^n+\frac{1}{12} \left(5-2
   \sqrt{6}\right)^{2 n}-\sqrt{\frac{2}{3}} \left(5+2
   \sqrt{6}\right)^n+\frac{1}{12} \left(5+2 \sqrt{6}\right)^{2 n}-\frac{1}{6}$
$b_n=\frac{1}{12} \left(5-2 \sqrt{6}\right)^{2 n}+\frac{1}{12} \left(5+2
   \sqrt{6}\right)^{2 n}+\frac{5}{6}$
$c_n=-\sqrt{\frac{2}{3}} \left(5-2 \sqrt{6}\right)^n+\frac{1}{12} \left(5-2
   \sqrt{6}\right)^{2 n}+\sqrt{\frac{2}{3}} \left(5+2
   \sqrt{6}\right)^n+\frac{1}{12} \left(5+2 \sqrt{6}\right)^{2 n}-\frac{1}{6}$.
For the first few integers $n$ we get the following triples :
$n=0$ => $\{0,1,0\}$
$n=1$ => $\{0,9,16\}$
$n=2$ => $\{720,801,880\}$
$n=3$ => $\{77616, 78409, 79200\}$.
