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
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