Multiple perfect squares

Note that if there is one solution, there are infinitely many; for if $(x,y,z)$ is a solution, so is $(n^2x,n^2y,n^2z)$ for positive integers $n$ .

Also, if $(x,y,z)$ is a solution without a square as common factor, then two of the numbers are multiples of four, and the other is one more than a multiple of four. Of the perfect squares, one is even and the other three are odd.

If we can find four distinct squares $a^2 < b^2 < c^2 < d^2$ such that $a^2 + b^2 + c^2 = 2d^2,$ then a solution is $x = (d - a)(d + a),\ \ y = (d-b)(d+b),\ \ z = (d-c)(d +c).$

I found $16^2 + 19^2 + 21^2 = 2\cdot 23^2,$ leading to the choices $x = 7\cdot 39 = 273;\ \ y = 4\cdot 42 = 168;\ \ z = 2\cdot 44 = 88.$ Then $x + y = 273 + 168 = 441 = 21^2; \\ y + z = 168 + 88 = 256 = 16^2; \\ z + x = 88 + 273 = 361 = 19^2; \\ x + y + z = 273 + 168 + 88 = 729 = 23^2.$

We are looking for distinct positive integers $a,b,c,d$ with $a^2 + b^2 + c^2 = 2d^2$ and $a,b,c < d$ . If we define $\alpha = \tfrac{a}{d}$ , $\beta = \tfrac{b}{d}$ , $\gamma = \tfrac{c}{d}$ , then $\alpha,\beta,\gamma$ are rationals with $0 < \alpha,\beta,\gamma < 1$ and $\alpha^2 + \beta^2 + \gamma^2 = 2$ .

Define $u,v \in \mathbb{Q}$ by the formulae $\alpha = 1 - u\gamma$ , $\beta = 1 - v\gamma$ , Then $\begin{aligned} (1 - u\gamma)^2 + (1 - b\gamma)^2 + \gamma^2 & = 2 \\ \gamma\big[(u^2 + v^2 + 1)\gamma - 2(u+v)\big] & = 0 \end{aligned}$ so that $\gamma = \frac{2(u+v)}{u^2 + v^2 + 1}$ Since $\gamma > 0$ and $\alpha,\beta < 1$ we deduce that $u,v > 0$ . Since $\gamma < 1$ we deduce that $2(u+v) < u^2 + v^2 +1$ , and hence that $(u-1)^2 + (v-1)^2 > 1$ . If $u,v < 1$ it is clear that $\alpha = 1 - u\gamma > 1 - u > 0 \hspace{1cm} \beta = 1 - v\gamma > 1 - v > 0$ Thus we obtain suitable values of $\alpha,\beta,\gamma$ , from which we can retrieve suitable values of $a,b,c,d$ , by choosing (nearly) any pair of rationals $u,v$ such that $0 < u < 1 \hspace{1cm} 0 < v < 1 \hspace{1cm} (u-1)^2 + (v-1)^2 > 1$ This is the region in the $uv$ -plane defined by the circle centre $(1,1)$ and radius $1$ , the $u$ -axis and the $v$ -axis.

If we choose $u \neq v$ , then $\alpha \neq \beta$ . The conditions for $\gamma$ to be distinct from $\alpha,\beta$ are more complex: $(u+v+1)^2 \neq 2(u^2+1) \hspace{2cm} (u+v+1)^2 \neq 2(v^2+1)$ but it is easy enough to choose $u,v$ in the above region which satisfy these extra conditions. Indeed, if we pick $u,v$ in the darker shaded region, we always obtain $0 < \alpha < \beta < \gamma < 1$ . The other five regions give the other possible orderings of $\alpha,\beta,\gamma$ .

For example, with $u = \tfrac13$ , $v = \tfrac{4}{21}$ we obtain $\alpha = \frac{16}{23} \hspace{1cm} \beta = \frac{19}{23} \hspace{1cm} \gamma = \frac{21}{23}$ and hence we can retrieve $a=16$ , $b=19$ , $c=21$ and $d=23$ .

Irritatingly, having done all the hard work to parametrize the solutions of the equation, I got the conditions on $u,v$ wrong initially, and made the wrong solution choice!

Diophantus of Alexandria set out a pattern for solving such problems - start out hopefully, see where the maths goes, and be perfectly content if it leads you to an answer! In this spirit let us call the three numbers we are looking for a,b and c and start with the observation that for any integer x, the three equations -

$a+b+c=(x+1)^2 \dots \boxed{1} \\ a+b=x^2 \dots \boxed{2} \\ b+c=(x-1)^2 \dots \boxed{3}$

guarantee that three of the conditions are automatically satisfied, and so the next step is to find x so that $a+c$ is also a square. Well, taking $2 \times \boxed{1}-\boxed{2}-\boxed{3}$ gives

$a+c=2(x+1)^2-x^2-(x-1)^2=6x+1$

So all we need to do is to find x, so that $6x+1$ is a square. There are an infinite number of possible values for $x$ which make $6x+1$ square, but to meet the full requirements of this problem we must check retrospectively that the integers a,b and c are indeed positive and distinct. For instance $x=20$ will do because $6x+1=121$ which is a square, and then substituting $x=20$ into our three numbered equations gives

$a+b+c=441 \\ a+b=400 \\ b+c=361$

which are easily solved to give distinct positive integers

$a=80 \\ b=320 \\ c=41$

which satisfy all the conditions of the problem.

One solution is x = 720, y = 801 and z = 880. We can check that 720 + 801 = 39^2, 801 + 880 = 41^2, 880 + 720=40^2 and 720 + 801 + 880 = 49^2.

As all that is needed to prove that a solution exists, is to find one example, it seems quickest to just start looking for one. Wrote a Python script in about 2 mins, just guessing a solution exists with $x, y, z \lt 1000$ :

from math import sqrt
def is_square(x, eps=1e-9):
    root = sqrt(x)
    return abs(int(root) - root) < eps

n = 1000
for x in range(1, n-2):
    for y in range(x+1, n-1):
        for z in range(y+1, n):
            if is_square(x+y) and is_square(y+z) and is_square(x+z) and is_square(x+y+z):
                print(x, y, z)

The question is, does there exists positive and distinct x, y, z such that x + y = a^2 , y + z = b^2 , x + z = c^2 and x + y + z = d^2 where a,b,c and d are integers. Solve this linear system to obtain that x = d^2 - b^2,
y = a^2 - d^2 + b^2, z = d^2 - a^2 provided that a^2 + b^2 + c^2 = 2d^2. Note that (11^2) + (19^2) + (20^2) = 2(21^2) so our restrictions are satisfied if a = 11 , b = 19 , c = 20 and d = 21. From this we obtain x = 80 , y = 41 , and z = 320. To verify, we see that x + y = 121, y + z = 361, x + z = 400, and x + y + z = 441.