Surfing on the integers

As in the inspiration question, we are looking for the smallest possible value of $d^2$ , where $0 < a < b < c < d$ are integers and $a^2 + b^2 + c^2 = 2d^2$ .

If $d$ is even, then $2d^2$ is a multiple of $8$ , and hence (since odd squares are congruent to $1$ modulo $8$ ), all of $a,b,c$ are even. Thus we could divide each of $a,b,c,d$ by $2$ , obtaining another solution with a smaller fourth value. Thus, for the smallest possible value of $d^2$ , $d$ will be odd. Since $0 < a < b < c < d$ , we deduce that $d \ge 5$ .

If $c < d\sqrt{\tfrac23}$ then $a^2 + b^2 + c^2 < 2d^2$ , which is impossible. Thus we must have $d > c > d\sqrt{\tfrac23}$ . There are a number of cases to consider:

• if $d=5$ then $c > 5\sqrt{\tfrac23}$ , and hence $c \ge 5$ . This is impossible.
• if $d=7$ then $c=6$ , so $a^2 + b^2 = 2d^2 - c^2 = 62 = 2\times31$ , which is impossible.
• if $d=9$ then $c=8$ , so $a^2 + b^2 = 2d^2 - c^2 = 98 = 7^2$ , and so $a=0,b=7$ , which is impossible.
• if $d=11$ then $c=9$ or $10$ . If $c=9$ then $a^2+b^2 = 161 = 7\times23$ , which is impossible. If $c=10$ then $a^2+b^2 = 142 = 2\times71$ , which is impossible.
• if $d=13$ then $c=11$ or $12$ . If $c=11$ then $a^2+b^2 = 217=7\times31$ , which is impossible. If $c=12$ then $a^2+b^2 = 194 = 2\times97 = (1+i)(1-i)(9+4i)(9-4i)$ , and hence $a=5,b=13$ , which is impossible.
• if $d=15$ then $c=13$ or $14$ . If $c=13$ then $a^2 + b^2 = 281 = 16^2 = 5^2$ , so $a=5,b=16$ , which is impossible. If $c=14$ then $a^2+b^2 = 254 = 2\times127$ , which is impossible.
• if $d=17$ then $14\le c \le16$ . If $c=14$ then $a^2+b^2 = 382 = 2\times191$ , which is impossible. If $c=15$ then $a^2+b^2 = 353 = 17^2 + 8^2$ , and so $a=8,b=17$ , which is impossible. If $c=16$ then $a^2+b^2 = 322 = 2\times7\times23$ , which Is impossible.
• if $d=19$ then $16 \le c \le 18$ . If $c=16$ then $a^2 + b^2 = 466 = 2\times233 =(1+i)(1-i)(13+8i)(13-8i)$ , and hence $a=5,b=21$ , which is impossible. If $c=17$ then $a^2+b^2=433 = 17^2 + 12^2$ , so $a=12,b=17$ , which is impossible. If $c=18$ then $a^2+b^2 =2^2 \times 97 = (1+i)^2(1-i)^2(9+4i)(9-4i)$ , which leads to $a=8,b=18$ , which is impossible.

On the other hand, $11^2 + 19^2 + 20^2 = 2\times21^2$ , and so the smallest value of $x+y+z$ is $21^2 = \boxed{441}$ ,

In the above I am using without comment the fact that an integer cannot be written as the sum of two nonzero squares if its prime factorisation contains a prime factor that is congruent to $3$ modulo $4$ and which has odd index, together with some basic ideas of factorisation in the Gaussian integers $\mathbb{Z}[ i ]$