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(A) Let the four integers in an increasing arithmetic progression be $a-3d, a-d, a+d, a+3d$ with common difference $2d > 0$. Alternatively,let the ter

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(A) Let the four integers in an increasing arithmetic progression be $a-3d, a-d, a+d, a+3d$ with common difference $2d > 0$. Alternatively,let the terms be $x, x+k, x+2k, x+3k$ where $k > 0$. We are given that one of these is the sum of the squares of the other three. Since the sequence is increasing,the largest term $x+3k$ must be the sum of the squares of the smaller three: $(x+3k) = x^2 + (x+k)^2 + (x+2k)^2$. Expanding this: $x+3k = x^2 + x^2 + 2xk + k^2 + x^2 + 4xk + 4k^2 = 3x^2 + 6xk + 5k^2$. Rearranging gives $3x^2 + (6k-1)x + (5k^2-3k) = 0$. For $x$ to be an integer,the discriminant $D = (6k-1)^2 - 4(3)(5k^2-3k)$ must be a perfect square. $D = 36k^2 - 12k + 1 - 60k^2 + 36k = -24k^2 + 24k + 1$. For $D \ge 0$,$-24k^2 + 24k + 1 \ge 0$. Testing $k=1$: $D = -24 + 24 + 1 = 1$,which is $1^2$. Then $x = \frac{-(6(1)-1) \pm 1}{2(3)} = \frac{-5 \pm 1}{6}$. $x = -1$ or $x = -2/3$ (not an integer). If $x = -1$ and $k = 1$,the terms are $-1, 0, 1, 2$. Check: $2 = (-1)^2 + 0^2 + 1^2 = 1 + 0 + 1 = 2$. This works. The terms are $-1, 0, 1, 2$. The common difference is $0 - (-1) = 1$.

Consider four distinct 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 common difference of the four numbers?

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