Please thx!

$\text{I'm going to call your roots }(r_{11},~r_{12},~r_{21},~r_{22})$

$r_{21}=r_{11}-1 \\ r_{22}=r_{12}-1$

$p_1=x^2+c x + a = (x-r_{11})(x-r_{12}) =x^2 -\left(r_{11}+r_{12}\right) x+r_{11} r_{12}\\ p_2 = x^2 +(2-r_{11}-r_{12})x + (1-r_{11}-r_{12}+r_{11}r_{12})$

$\text{the coefficient }a \text{ appears in both polynomials so we can solve for the roots}\\ r_{11}r_{12} = 2-r_{11}-r_{12} \\ \text{Solving this we find that there are two possibilities}\\ (r_{11},~r_{12}) = (-2,~-4) ~\text{or}\\ (r_{11},~r_{12}) = (0,~2)$

$\text{In the first case we then have}\\ (r_{21},~r_{22}) = (-3,-5) \\ p_1(x) = (x+2)(x+4) = x^2 + 6x+8 \Rightarrow a=8,~c=6\\ p_2(x) = (x+3)(x+5) = x^2 + 8x + 15 \Rightarrow b=15 \\ a+b+c = 8 + 15 + 6 = 29$

$\text{In the second case we have}\\ (r_{21},~r_{22}) = (-1,1)\\ p_1(x) =(x-0)(x-2) = x^2 - 2x \Rightarrow a=0,~c=-2\\ p_2(x) = (x+1)(x-1) = x^2 - 1 \Rightarrow b=-1\\ a+b+c = 0-1-2 = -3$

Someone please double check all of this.

I got to the same result, but by a different route.

To begin with, remember that if y = f(x), then replacing x by x + k shifts the graph of f(x) a distance k to the left.

So, since the roots of the second quadratic are each 1 less than the roots of the first quadratic, it follows that the second quadratic must be (x + 1)^2 + c(x + 1) + a.

Collecting up terms and equating coefficients, c + 2 = a and 1 + c + a = b. 

A further requirement is that the roots are to be integers, so, from the first quadratic x^2 + cx + (c + 2) = 0,

$\displaystyle x = \frac{-c \pm\sqrt{c^{2}-4(c+2)}}{2}\; .$

The expression under the root sign will need to be the square of an integer,  so, completing the square,

$\displaystyle c^{2}-4c-8=c^{2}-4c+4-12=(c-2)^{2}-12=m^{2}\; ,$    m an integer.

The only two squares differing by 12 are 4 and 16, so c = 6 or c = -2.

If c = 6, then a = 8, b = 15, so a + b + c = 29.

If c = -2, then a = 0, b = -1, so a + b + c = -3.

Tiggsy