Solving equations (part 2)

How can the numbers 7,13,20,27 be in A.P.?

$(a,b,c,d)=(6,36,66,96)$ is another valid solution, so there are (at least?) $5$ such consistent systems (not that this changes the answer).

My workings were pretty messy - lots of Vieta, a bit too much reliance on Wolfram|Alpha...would you be able to post a full solution?

I got the point. Thanks. May be there are more than five solutions. I'm checking it now.

From the given equations, it can be shown that $a^4 - 6a^2 b + 8ac + 3b^2 - 6d = 0.$ Then using the arithmetic progression condition, this reduces to $a^4 - 6a^2 b - 8a^2 + 16ab + 3b^2 + 12a - 18b = 0.$

Here's a solution with a bit more detail:

We have $a^3=x^3+y^3+z^3+3(xy^2+yz^2+zx^2+x^2y+y^2z+z^2x) +6xyz$ , so $a^3-3ab+2c-6xyz=0 \text{ (I)}$

Similarly we have $a^4 =x^4+y^4+z^4+4(xy^3+yz^3+zx^3+x^3y+y^3z+z^3x)+6(x^2y^2+y^2z^2+z^2x^2) + 12axyz$

$=d+4(ac-d)+3(b^2-d) + 12axyz$ , so

$a^4-4ac-3b^2+6d-12axyz=0 \text{ (II)}$

Eliminate $xyz$ by multiplying $(I)$ by $2a$ and then subtract $(II)$ from it:

$a^4-6a^2b+8ac+3b^2-6d=0$

Using the arithmetic progression, now substitute $c=2b-a$ and $d=3b-2a$ and simplify to

$a^4 - 6a^2b - 8a^2 + 16ab + 3b^2 + 12a - 18b = 0$

We can view this as a quadratic in b:

$(3)b^2  +( -6a^2+16a -18)b +(a^4 -8a^2+12a) = 0$

with solutions $b=\frac{3a^2-8a +9 \pm \sqrt{ (-3a^2+8a -9)^2 - 3(a^4 -8a^2+12a )}} {3}$

in which the discriminant simplifies to

$D = 6a^4 -48a^3+142a^2-180a+81$ D needs to be a perfect square, so we can check different values of a, and find that D is a square for $a \in \{0,1,2,3,6,7\}$ . This is not exhaustive, but by checking quadratic residues mod p for a prime p, I could exclude many candidates at a time. Doing so for all primes $p \leq 150$ , I sieved out all other candidate values $a\leq 1,000,000$ .

Checking the found values for $a$ , we find 5 solutions for $(a,b)$ :

• Case $a=0$ : $b= \frac{9 \pm 9}{3}$ , of which $b=6$ is a possible solution satisfying $b>a$ .
• Case $a=1$ : $b= \frac{4 \pm 1}{3}$ , of which $b=1$ is the only integer solution.
• Case $a=2$ : $D=1, b=\frac{12-16+9 \pm 1}{3}$ , with integer solution $b=2$ . Because of the condition $b \gt a$ , this is no valid solution.
• Case $a=3$ : $D=3^2, b=\frac{27-24+9  \pm 3}{3}$ , with solutions $b=3$ or $b=5$ , of which only $b=5$ satisfies $b>a$ .
• Case $a=6$ : $D=39^2, b=\frac{108-48+9 \pm 39}{3}$ , with integer solutions $b=10$ and $b=36$ .
• Case $a=7$ : $D=61^2, b=\frac{147-56+9 \pm 61}{3}$ , with integer solution $b=13$ .
• $c$ and $d$ follow directly from $a$ and $b$ .
• We have not proved yet that to each a quadruple $(a,b,c,d)$ indeed one or more triples $(x, y, z)$ exists, but Wolfram alpha could find values for each of the above cases (even providing an analytic solution for the case $a=7$ ).

Summarized we found solutions $(a,b,c,d) \in \{(0,6,12,18),(3,5,7,9),(6,10,14,18),(6,36,66,96),(7,13,19,25)\}$