Solutions are given here for finding sets of three numbers such that the sum of any two of them is a perfect square. Can you generalize the methods to find sets of four numbers with this property?

Hyeyoun Chung from St Paul's Girls' School noticed that if the square numbers chosen are: $x^2$, $(x+1{)}^2$ and $(x+3{)}^2$ and the corresponding set of three integers are $a,\hspace{1em}b$ and $c$, with $a<b<c$ and:

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{a+b}& =x^2\\ \multicolumn{1}{c}{a+c}& =(x+1{)}^2\\ \multicolumn{1}{c}{b+c}& =(x+3{)}^2\end{array}.}$

then these these three equations can be solved to give:

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{c}& =\frac{x^2+8x+10}{2}\\ \multicolumn{1}{c}{b}& =\frac{x^2+4x+8}{2}\\ \multicolumn{1}{c}{a}& =\frac{x^2-4x-8}{2}\end{array}.}$

Therefore, from a single square number, $x^2$, three integers $a$, $b$ and $c$ can be found with values such that the sum of any pair of these numbers is a perfect square. The only problem with this system is, however, that the square number must be even, otherwise $a$, $b$ and $c$ will not be integers.

Yatir Halevi tackled the problem as follows:

We are given 3 square numbers: $a^2,\hspace{1em}{b}^2$ and $c^2$, and we have to find a general formula for finding 3 integers that are 'Pair Squares' such that the sum of any pair is a square number, that is: $x+y=a^2$, $y+z=b^2$ and $x+z=c^2$.

Solving for $x,\hspace{1em}y$ and $z$ we get:

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{a^2-b^2}& =x-z\\ \multicolumn{1}{c}{c^2}& =x+z\\ \multicolumn{1}{c}{2x}& =a^2+c^2-b^2\\ \multicolumn{1}{c}{x}& =(a^2-b^2+c^2)/2\end{array}}$

By the same methods we get:

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{y}& =(a^2+b^2-c^2)/2\\ \multicolumn{1}{c}{z}& =(-a^2+b^2+c^2)/2.\end{array}}$

In order for $x,\hspace{1em}y$ and $z$ to be integers; either $a,\hspace{1em}b$ and $c$ are all even or two are odd and the other is even because if an integer $n$ is even or odd so is its square. This is proved as follows: if $n$ is odd then it can be written $n=2m+1$, so $n^2=(2m+1{)}^2=4m^2+4m+1=2(2m^2+2m)+1=2q+1$ which is odd. If $n$ is even then $n=2m$, so $n^2=(2m{)}^2=4m^2=2(m^2)$ which is even.

For the first case we can generate Pair Squares by choosing any three even squares, for example 4, 16 and 36 and the formula gives:

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{x}& =(4-16+36)/2=12\\ \multicolumn{1}{c}{y}& =(4+16-36)/2=-8\\ \multicolumn{1}{c}{z}& =-4+16+36)/2=24.\end{array}}$

This is a set of pair squares because $12-8=4$, $-8+24=16$ and $24+12=36$.

For the second case we can generate Pair Squares by choosing any two odd squares together with one even square, for example 9, 1 and 4 and the formula gives:

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{x}& =(9-1+4)/2=6\\ \multicolumn{1}{c}{y}& =(9+1-4)/2=3\\ \multicolumn{1}{c}{z}& =-9+1+4)/2=-2.\end{array}}$

Checking that this is a set of pair squares we have $6+3=9,\hspace{1em}3-2=1,\hspace{1em}6-2=4$.

As for four integers that are pair squares: take an even square $a^2$ and halve it and you have $b=a^2/2$ giving $\{b,\hspace{1em}b,\hspace{1em}b,\hspace{1em}b\}$ for example $\{8,8,8,8\}$ or $\{8,8,8,17\}$.

Saul Foresta used a similar method and gave the following example: $x=54$, $y=10$ and $z=90$