When the $a^2+b+c+d,b^2+a+c+d,c^2+a+b+d,d^2+a+b+c$ are all perfect squares?

Question

Find all $a,b,c,d\in \mathbb{Z}^+$, which $a^2+b+c+d,b^2+a+c+d,c^2+a+b+d,d^2+a+b+c$ are all perfect squares.

I found $(1,1,1,1)$, but I can't find more.

Is $a=b=c=d$ true?

Answer 1

The solutions are $(11, 11, 6, 6)$, $(96, 96, 57, 40)$, and $(x, x, x, 1)$, where $3x + 1$ is a square.

Without loss of generality, $a \ge b \ge c \ge d$, since it is symmetric (not just cyclic). Then$$a^2 < a^2 + b + c + d \le a^2 + 4a < (a+2)^2 \implies a^2 + b + c + d = (a+1)^2.$$Thus, $b + c + d = 2a + 1$. Then$$b \ge {1\over3}(2a + 1) \implies a \le {3\over2}b,$$so$$b^2 < b^2 + c + d + a \le b^2 + {7\over2}b < (b+2)^2 \implies b^2 + c + d + a = (b+1)^2.$$Thus, $b + c + d = 2a + 1$ and $c + d + a = 2b + 1$, which implies $a = b$, thus $c + d = a + 1$.

Consequently,$$c \ge{1\over2}(2a+1) \implies a < 2c,$$so$$(c+1)^2 \le c^2 + 2a + d \le c^2 + 5c < (c+3)^2,$$and hence we have two cases.

Case 1. If $c^2 + 2a + d = (c+1)^2$, we must have $a = b = c$ and $d = 1$. Thus, we find that $(a, b, c, d) = (x, x, x, 1)$ where $x$ is such that $3x + 1$ is a square. This is one set of solutions.

Case 2. If $c^2 + 2a + d = (c + 2)^2$, so $2a + d = 3c + 4$. Combined with $c + d = 2a + 1$, we derive that $a = (5/2)d - 4$ and $c = (3/2)d - 3$. So$$(d+1)^2 \le d^2 + 2a + c = d^2 + {{13}\over2}d - 11 < (d+4)^2.$$So setting this equal to $(d+1)^2$, $(d+2)^2$, $(d+3)^2$, we find the integer solutions $d = 6$ and $d = 40$. This gives the solutions $(11, 11, 6, 6)$ and $(96, 96, 57, 40)$.

Hence, the solutions are $(11, 11, 6, 6)$, $(96, 96, 57, 40)$, and $(x, x, x, 1)$, where $3x + 1$ is a square.

Answer 2

Since the expressions are symmetric, then we can set $a\leq b\leq c\leq d$ without loss of generality. There is an infinite number of positive integer solutions if we assume $a=1$ and $b=c=d=n$ which yields,

$$\begin{aligned} a^2+b+c+d &= 3n+1\\ a+b^2+c+d &= (n+1)^2\\ a+b+c^2+d &= (n+1)^2\\ a+b+c+d^2 &= (n+1)^2 \end{aligned}\tag1$$

and it is easy to find solve $3n+1 = y^2$.

Excluding this infinite family, it seems the only solutions with all variables $1\leq a\leq b\leq c\leq d < 100$ are,

$$a,b,c,d = 6,6,11,11$$

$$a,b,c,d = 40,57,96,96$$

though I am not sure of this result. (Anyone can verify it?)

Answer 3

Do brute force search. Here is a piece of Python code that should do the job

from math import sqrt, floor

_is_whole = lambda x: floor(x) == x
_is_ps = lambda n: _is_whole(sqrt(n))

min_int = 1
max_int = 150
for a in range(min_int, max_int + 1):
    for b in range(a, max_int + 1):
        for c in range(b, max_int + 1):
            for d in range(c, max_int + 1):
                for x in [a ** 2 + b + c + d, a + b ** 2 + c + d,
                          a + b + c ** 2 + d, a + b + c + d ** 2]:
                    if not _is_ps(x):
                        break
                else:
                    print a, b, c, d

When run, it should produce the following output:

1 1 1 1

1 5 5 5

1 8 8 8

1 16 16 16

1 21 21 21

1 33 33 33

1 40 40 40

1 56 56 56

1 65 65 65

1 85 85 85

1 96 96 96

1 120 120 120

1 133 133 133

6 6 11 11

40 57 96 96

Answer 4

I think this system of Diophantine equations.

$$\left\{\begin{aligned}&A^2+(B+C+D)Q=X^2\\&B^2+(A+C+D)Q=Y^2\\&C^2+(A+B+D)Q=Z^2\\&D^2+(A+B+C)Q=W^2\end{aligned}\right.$$

It is better to solve in General. This notation allows to find solutions for any values $Q$. Solution we write better.

$$A=(2b+c)(2f+k)ckt^2p^2+2((2b+c)ct^2f^2+2(b^2t^2+2bca^2)fk+(b^2t^2-a^2c^2)k^2)ps+$$

$$+((2b+c)ct^2f^2+2(b^2t^2-a^2c^2)fk+(b^2t^2-2(b+c)ca^2)k^2)s^2$$

$$***$$

$$B=(2a+t)(2f+k)tkc^2p^2+2((2a+t)tc^2f^2+2(a^2c^2+2atb^2)fk+(a^2c^2-b^2t^2)k^2)ps+$$

$$+((2a+t)tc^2f^2-2(b^2t^2-a^2c^2)fk+(a^2c^2-2(a+t)tb^2)k^2)s^2$$

$$***$$

$$C=(2b+c)(2a+t)ctk^2p^2+2((4ab-tc)tcf^2+((2a+t)tb^2+(2b+c)ca^2)k^2)ps+$$

$$+(((2a+t)tb^2+(2b+c)ca^2)k^2-2(ac+bt+tc)tcf^2)s^2$$

$$***$$

$$D=2((4ab-tc)f-(ac+bt+tc)k)tckp^2+$$

$$+((2b+c)(2a+t)tcf^2+((2b+c)ca^2+(2a+t)tb^2)(2f+k)k)s^2$$

$$***$$

$$Q=tcsk(4((tc-4ab)f+(ac+bt+tc)k)p+$$

$$+(4(ac+bt+tc)f+(4ab+4ac+4bt+3tc)k)s)$$

$$***$$

$$X=(2b+c)(2f+k)ckt^2p^2+$$

$$+2((2b+c)ct^2f^2+2(t^2b^2+(ct-2ab)ac)fk+((ac+2bt+2ct)ac+t^2b^2)k^2)ps+$$

$$+((2b+c)ct^2f^2+2((ac+2bt+2ct)ac+t^2b^2)fk+$$

$$+(2(b+c)ca^2+(4b+3c)atc+t^2b^2)k^2)s^2$$

$$***$$

$$Y=(2a+t)(2f+k)tkc^2p^2+$$

$$+2((2a+t)tc^2f^2+2(c^2a^2+(tc-2ab)bt)fk+(a^2c^2+(2ac+bt+2tc)bt)k^2)ps+$$

$$+((2a+t)tc^2f^2+2(a^2c^2+(2ac+bt+2ct)bt)fk+(2(a+t)tb^2+(4a+3t)tbc+a^2c^2)k^2)s^2$$

$$***$$

$$Z=(2a+t)(2b+c)tck^2p^2+$$

$$+2((tc-4ab)tcf^2+2(ac+bt+tc)tcfk+((2a+t)tb^2+(2b+c)ca^2)k^2)ps+$$

$$+(2(ac+bt+tc)tcf^2+(4ab+4bt+4ac+3tc)tcfk+((2a+t)tb^2+(2b+c)ca^2)k^2)s^2$$

$$***$$

$$W=2((tc-4ab)f+(ac+bt+tc)k)tckp^2+$$

$$+(4(ac+bt+tc)f+(4ab+4ac+4bt+3tc)k)tckps+$$

$$+((2b+c)(2a+t)tcf^2+((2b+c)ca^2+(2a+t)tb^2)(2f+k)k)s^2$$

$a,t,b,c,f,k,p,s - $ integers which we can ask.

If we want to find out what the odds should be for example $Q=1 , 2 , 3 ....$ . Just substitute the formula and solve the equation. Everything will be reduced to the factorization.