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
$$a^{2}+b^{2}+c^{2} =d^{2} + d^{2} + d^{2} + (d-1)^{2}$$
 A: This isn't a full answer, but noting that $$2a+2b+2c=6d$$ first subtract this from the second equation to obtain $$(a-1)^2+(b-1)^2+(c-1)^2=4(d-1)^2$$
Then set $A=a-1, B=b-1, C=c-1, D=d-1$ to get the system$$A^2+B^2+C^2=4D^2$$
and $$A+B+C=3D$$
So for given $D$ the solutions lie on a sphere of radius $2D$ centred at the origin.

@Will Jagy has completed this to a full solution in the comments below - noting that if we don't have $A=B=C=D=0$ then modulo $4$ considerations give a contradiction.
One motivation for looking for something like this is the existence of the solution $a=b=c=d=1$: how would one encode that in an equation or (in different circumstances) produce a factor which could be divided out to simplify the search for other solutions.
Will should get credit for the solution here.

In response to request to complete this rather than leaving the finish in comments:
Now suppose we have a non-zero solution. If $A,B,C,D$ are all even we can divide through by $2$ to get another solution (because the two equations we now have are homogeneous). So if there is a non-zero solution, there is a non-zero solution where at least one of $A,B,C,D$ is odd.
However the sum of three squares can only be divisible by $4$ if it is the sum of three even squares (any odd square is $\equiv 1 \bmod 4$ - indeed $\equiv 1 \bmod 8$), so $A,B,C$ must all be even to satisfy the first equation. Then the second equation tells us that $D$ is even. And this is a contradiction to the fact that we can find a solution with at least one odd number. Hence there is no non-zero solution.
A: 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.
A: $a+b+c=3d, a^2+b^2+c^2=4d^2-2d+1$
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.
