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(A) Let the four consecutive integers be $a-3d, a-d, a+d, a+3d$. However,since they are consecutive integers,the common difference must be $1$. Let th

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$2$

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(A) Let the four consecutive integers be $a-3d, a-d, a+d, a+3d$. However,since they are consecutive integers,the common difference must be $1$. Let the integers be $x, x+1, x+2, x+3$.We are given that one of these integers is the sum of the squares of the other three.Case $1$: $x+3 = x^2 + (x+1)^2 + (x+2)^2$$x+3 = x^2 + x^2 + 2x + 1 + x^2 + 4x + 4$$3x^2 + 5x + 2 = 0$$(3x+2)(x+1) = 0$$x = -1$ or $x = -2/3$ (not an integer).If $x = -1$,the integers are $-1, 0, 1, 2$.Check: $2 = (-1)^2 + 0^2 + 1^2 = 1 + 0 + 1 = 2$. This holds true.The sum of these integers is $(-1) + 0 + 1 + 2 = 2$.

Consider four consecutive integers in an increasing arithmetic progression. One of these integers is equal to the sum of the squares of the other three. What is the sum of all the numbers?

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