<s>[INST] Given, $2\le aGiven, $2\le a<b<c<d\le 16$ and $(d-1)^2=(a-1)^2+(b-1)^2+(c-1)^2$,$(d+1)^2+(a+1)^2=(b+1)^2+(c+1)^2$. Find all $a,b,c,d$  
My work:
If $(a,b,c,d)\in \mathbb{N}$, for that I could show this
$$(d-1)^2=(a-1)^2+(b-1)^2+(c-1)^2\implies d^2-2d=(a^2+b^2+c^2)-2(a+b+c)+2$$
Now, From second equation we get,
$$d^2+2d+a^2+2b+2=b^2+c^2+2b+2c+2$$
Now, manipulating this equation to look like the above equation and substitution gives,
$$4d+2(a^2+1)=4(b+c)\implies a^2\equiv -1 \mod2$$
From this we can get that $a$ is odd and $b+c>d$.
Now what am I supposed to do? I could not approach this problem for $(a,b,c,d)\in \mathbb{R}\setminus \mathbb{N}$. Please help.
 [/INST] Start as you did (i.e. by expanding, simplifying, and “equating” the two given equalities), then solve for $a$ to obtain
$$a = \sqrt{2(b+c-d)-1}.$$
Note that $b+c \le 29$, so $b+c-d \le 24$, so $2(b+c-d)-1 \le 47$. Since $2(b+c-d)-1$ must be an odd square, and $a \ge 2$, we conclude $2(b+c-d)-1 \in \{9,25\}$. Hence $a=3$ or $a=5$. The latter is easily eliminated, which quickly leads to $(a,b,c,d)=(3,7,10,12)$ or $(a,b,c,d)=(3,6,15,16)$.
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