<s>[INST] When the $a^2+b+c+d,b^2+a+c+d,c^2+a+b+d,d^2+a+b+c$ are all perfect squares? Find all $a,b,c,d\in \mathbb{Z}^+$, which $a^2+b+c+d,b^2+a+c+d,c^2+a+b+d,d^2+a+b+c$ are all perfect squares.
I found $(1,1,1,1)$, but I can't find more.
Is $a=b=c=d$ true?
 [/INST] The solutions are $(11, 11, 6, 6)$, $(96, 96, 57, 40)$, and $(x, x, x, 1)$, where $3x + 1$ is a square.
Without loss of generality, $a \ge b \ge c \ge d$, since it is symmetric (not just cyclic). Then$$a^2 < a^2 + b + c + d \le a^2 + 4a < (a+2)^2 \implies a^2 + b + c + d = (a+1)^2.$$Thus, $b + c + d = 2a + 1$. Then$$b \ge {1\over3}(2a + 1) \implies a \le {3\over2}b,$$so$$b^2 < b^2 + c + d + a \le b^2 + {7\over2}b < (b+2)^2 \implies b^2 + c + d + a = (b+1)^2.$$Thus, $b + c + d = 2a + 1$ and $c + d + a = 2b + 1$, which implies $a = b$, thus $c + d = a + 1$.
Consequently,$$c \ge{1\over2}(2a+1) \implies a < 2c,$$so$$(c+1)^2 \le c^2 + 2a + d \le c^2 + 5c < (c+3)^2,$$and hence we have two cases.
Case 1. If $c^2 + 2a + d = (c+1)^2$, we must have $a = b = c$ and $d = 1$. Thus, we find that $(a, b, c, d) = (x, x, x, 1)$ where $x$ is such that $3x + 1$ is a square. This is one set of solutions.
Case 2. If $c^2 + 2a + d = (c + 2)^2$, so $2a + d = 3c + 4$. Combined with $c + d = 2a + 1$, we derive that $a = (5/2)d - 4$ and $c = (3/2)d - 3$. So$$(d+1)^2 \le d^2 + 2a + c = d^2 + {{13}\over2}d - 11 < (d+4)^2.$$So setting this equal to $(d+1)^2$, $(d+2)^2$, $(d+3)^2$, we find the integer solutions $d = 6$ and $d = 40$. This gives the solutions $(11, 11, 6, 6)$ and $(96, 96, 57, 40)$.
Hence, the solutions are $(11, 11, 6, 6)$, $(96, 96, 57, 40)$, and $(x, x, x, 1)$, where $3x + 1$ is a square.
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