<s>[INST] Find the integer solution: $a+b+c=3d$, $\: a^{2} + b^{2} + c^{2}= 4d^{2}-2d+1$ Find  the integer solutions:
$$a+b+c=3d$$
$$a^{2} + b^{2} + c^{2}= 4d^{2}-2d+1$$

Attempt:
Notice that $a=b=c=d=1$ is a solution.

Other facts:
Notice that $a^{2} + b^{2} + c^{2} > 0$, so $4d^{2}-2d+1>0$. Notice that $4d^{2}-2d+1$ has negative discriminant: $D = -12$, and so it is always one sign, which is positive in this case. So we cannot narrow $d$-solution this way.
Next, since $(a+b+c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab+ac+bc)$, we get
$$ 9d^{2} - 2(ab+ac+bc) = 4d^{2}-2d+1  $$
$$ 5d^{2} + 2d -1 = 2(ab+ac+bc) $$
$5d^{2} + 2d - 1 = 0$ has solutions:
$$ d_{1,2} = \frac{-2 \pm \sqrt{24}}{10} = \frac{-1 \pm \sqrt{6}}{5}$$
and when $d > (-1 + \sqrt{6})/5$, OR, $d < (-1 - \sqrt{6})/5$ we have
$5d^{2}+2d-1 > 0$. And when $d$ is between the 2 roots we have $5d^{2}+2d-1<0$.
Then also nocitce that $(a+b+c)d= 3d^{2}$, and so
$$a^{2}+b^{2}+c^{2}-(a+b+c)d = (d-1)^{2}$$
Now if $d \ne 1$ we must have
$$ a^{2}+b^{2}+c^{2}-(a+b+c)d > 0$$
Another fact:
$$d = \frac{a+b+c}{3} \ge (abc)^{1/3} $$
by AM-GM.
Also notice $d < \max(a,b,c)$, because
$$a^{2}+b^{2}+c^{2} =d^{2} + d^{2} + d^{2} + (d-1)^{2}$$
 [/INST] Your equation has no non-zero solutions.
Lemma. The Diophantine equation $3x^2+y^2=2z^2$ has no non-zero solutions in integers. Proof. Let $x_0$ is smallest positive integer such that there are integers $y_0$ and $z_0$ such that  $3x_0^2+y_0^2=2z_0^2$. If $y_0 \not \equiv 0 \pmod 3$ then $z_0 \not \equiv 0 \pmod 3$. Then $y_0^2 \equiv 1 \pmod 3$ and $2z_0^2 \equiv 2 \pmod 3$. Then $2z_0^2-y_0^2 \equiv 1 \pmod 3$. But $2z_0^2-y_0^2=3x_0^2 \equiv 0 \pmod 3$. Therefore $y_0 \equiv 0 \pmod 3$. Then $z_0 \equiv 0 \pmod 3$. Let $y_0=3k$, $z_0=3n$. Then $3x_0^2+9k^2=18n^2$. Then $x_0^2+3k^2=6n^2$. Then $x_0 \equiv 0 \pmod 3$. Let $x_0=3p$. Then $9p^2+3k^2=6n^2$. Then $3p^2+k^2=2n^2$.We have come to a contradiction because $0<p<x_0$ and $p$ is integer solution of $3x^2+y^2=2z^2$.
Now solve your equation. Let $z=a+c$, $y=c-a$. Then $c=\frac{z+y}{2}$, $a=\frac{z-y}{2}$. Then from $a+b+c=3d$ we have $z+b=3d$. Then $b=3d-z$. Then $\left( \frac{z-y}{2}\right)^2+\left( \frac{z+y}{2}\right)^2+(3d-z)^2=4d^2-2d+1$. Then $z^2+y^2+2(3d-z)^2=8d^2-4d+2$. Then $3z^2-12zd+y^2=-10d^2-4d+2$. Then $3(z-2d)^2+y^2=2d^2-4d+2$. Then $3(z-2d)^2+y^2=2(d-1)^2$.The last equation has no non-zero solutions by the lemma.
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