# Solve the system of diophantine equations: $xz=a(a+b)$; $xt+yz=2ab^2$; $yt=b^4$.

I am trying to solve the following system of diophantine equations for $x,y,z,t$ where $\gcd(a,b)=1$ are coprime: $$xz=a(a+b)$$ $$xt+yz=2ab^2$$ $$yt=b^4$$ I noticed that $x,z$ share a common factor and $y,t$ have the same set of primes. It's fairly easy to make an assumption about those integers' prime factorization. However, it gets confusing when I use the values of those integers to satisfy the second equation. Any input will be appreciated.

• It is worth noting that if $xy\neq0$ then $-ab$ must be a square. – Servaes Nov 1 '17 at 1:25

If prime $p$ divides $a$, it can't divide $b$ and therefore (by the third equation) can't divide $y$ or $t$. By the first equation, it must divide $x$ or $z$. From the second equation, if it divides one of $x$ and $z$ it must divide the other. Moreover, from the second equation the highest powers of $p$ dividing $x$ and $z$ (say $p^k$) must be equal, and from the first equation $p^{2k}$ is the highest power of $p$ dividing $a$. Thus $a$ must be a square. Any other factors of $x$ and $z$ are factors of $a+b$.
Similarly, if prime $q$ divides $b$, then $q$ can't divide $a$ or $a+b$ or $x$ or $z$, and since it divides one of $y$ and $t$ if divides both. If $q^k$ and $q^m$ are the highest powers of $q$ dividing $y$ and $t$ respectively, then $q^{(k+m)/4}$ is the highest power dividing $b$, and $q^{(k+m)/2}$ is the highest power dividing $2ab^2$ unless $q = 2$. Now we get a contradiction in the second equation if $k \ne m$. Thus we must have $k = m$. Since this is true for all $q$, we conclude that $y = t = b^2$.
Thus we can write \eqalign{a &= A^2\cr x &= A u\cr z &= A v\cr b &= u v - A^2\cr y &= t = b^2\cr u + v &= 2 A b^2 = 2 A (uv - A^2)^2\cr} Solving this last equation for $v$, we find that $8 A^3 u + 8 A u^3 + 1$ must be a square.
Some examples of solutions are $$\matrix{A = 2 & u = 1 & v = 3 & a = 4 & b = -1 & x = 2 & y = t = 1 & z = 6\cr A = 1 & u = 2 & v = 0 & a = 1 & b = -1 & x = 2 & y = t = 1 & z = 0\cr A = 3 & u = 2 & v = 4 & a = 9 & b = -1 & x = 6 & y = t = 1 & z = 12\cr A = 2 & u = 3 & v = 1 & a = 4 & b = -1 & x = 6 & y = t = 1 & z = 2\cr A = 4 & u = 3 & v = 5 & a = 16 & b = -1 & x = 12 & y = t = 1 & z = 20\cr A = 3 & u = 4 & v = 2 & a = 9 & b = -1 & x = 12 & y = t = 1 & z = 6\cr A = 5 & u = 4 & v = 6 & a = 25 & b = -1 & x = 20 & y = t = 1 & z = 30\cr A = 4 & u = 5 & v = 3 & a = 16 & b = -1 & x = 20 & y = t = 1 & z = 12\cr A = 6 & u = 5 & v = 7 & a = 36 & b = -1 & x = 30 & y = t = 1 & z = 42\cr A = 5 & u = 6 & v = 4 & a = 25 & b = -1 & x = 30 & y = t = 1 & z = 20\cr }$$
Hmmm: all solutions I've found have $b=-1$.
• Note that $x=a\pm\sqrt{-ab}$ and $z=a\mp\sqrt{-ab}$, so $b$ must be the negative of a square. Conversely if $a=A^2$ and $b=-B^2$ then there are only two solutions: $x=A(A\pm B)$, $z=A(A\mp B)$, $y=t=B^4$. – Servaes Nov 1 '17 at 1:32