2016/027) Solve for integers $a,b,c$ given $a^2+b^2+c^2 + a + b+ c = 1$

Multiplying by 4 and adding 3 on both sides we get
$(4a^2+4a+1) + (4b^2 + 4 b+ 1) + (4c^2 + 4c+1) = 7$
or $(2a+1)^2 + (2b+1) ^2 + (2c+1)^2   = 7$
as for n odd $n^2 = 1 \, mod \, 8$
we have $LHS = 3\, mod \, 8$ and $RHS = 7 \,mod \, 8$ so no solution