Find the integer solution: $a+b+c=3d$, $\: a^{2} + b^{2} + c^{2}= 4d^{2}-2d+1$

Question

Find the integer solutions: $$a+b+c=3d$$ $$a^{2} + b^{2} + c^{2}= 4d^{2}-2d+1$$

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Attempt:

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

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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

$$a^{2}+b^{2}+c^{2} =d^{2} + d^{2} + d^{2} + (d-1)^{2}$$

Answer 1

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.

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@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.

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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.

Answer 2

$a+b+c=3d, a^2+b^2+c^2=4d^2-2d+1$

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.

Answer 3

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.