<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] $a+b+c=3d, a^2+b^2+c^2=4d^2-2d+1$
Just set $(a, b, c)=(k_1d+l_1, k_2d+l_2, k_3d+l_3)$, since if we have quadratic polynomials for $a, b, c$, the 4th degree polynomials have to be in $a^2+b^2+c^2$.
$4d^2-2d+1=(k_1^2+k_2^2+k_3^2)d^2+2(k_1l_1+k_2l_2+k_3l_3)d+(l_1^2+l_2^2+l_3^2).$
$k_1^2+k_2^2+k_3^2=4, k_1l_1+k_2l_2+k_3l_3=-1, l_1^2+l_2^2+l_3^2=1.$
$k_1+k_2+k_3=3, l_1+l_2+l_3=0.$
By solving this, you can get:
$(k_1, k_2, k_3, l_1, l_2, l_3) \\=\Bigg(k_1, -\dfrac {\pm\sqrt{-3k_1^2+6k_1-1}+k_1-3} {2}, \dfrac {\pm\sqrt{-3k_1^2+6k_1-1}-k_1+3} {2}, 1-k_1, \dfrac {\pm\sqrt{-3k_1^2+6k_1-1}+k_1-1} {2}, -\dfrac {\pm\sqrt{-3k_1^2+6k_1-1}-k_1+1} {2}\Bigg)$.
So,$a=k_1d+(1-k_1) \\ b=-\dfrac {\pm\sqrt{-3k_1^2+6k_1-1}+k_1-3} {2}d+\dfrac {\pm \sqrt{-3k_1^2+6k_1-1}+k_1-1} {2} \\ c=\dfrac {\pm \sqrt{-3k_1^2+6k_1-1}-k_1+3} {2}d-\dfrac {\sqrt{-3k_1^2+6k_1-1}-k_1+1} {2}$.
For instance, $a=d, b=\dfrac {2-\sqrt{2}} {2}d+\dfrac {1} {\sqrt{2}}, c=\dfrac {2+\sqrt{2}}{2}d-\dfrac {1} {\sqrt{2}}.$
Since $a, b, c,  d$ are all integers, we can't find the integers in this solution.
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