A number theory problem by Vitor Santos

SInce $xt + yz = 1$ , $x$ and $y$ are coprime. Now $xz - 2yt = 3(xt + yz)$ , so we see that $x(z-3t) \; = \; y(3z+2t)$ and hence, since $x$ and $y$ are coprime, we have $3z + 2t \; = \; ex \hspace{2cm} z - 3t \; = \; ey$ for some integer $e$ . Since $xt + yz = 1$ we deduce that $e \; = \; z^2 + 2t^2$ There are two cases to consider:

• If $11$ divides $e$ , we have $e = 11f$ for some integer, and hence $z = f(3x+2y)$ , $t = f(x-3y)$ , and $1 = xt + yz = f(x^2 + 2y^2)$ . This means that $f=1$ , $x = \pm1$ and $y=0$ , and hence $z = 3x$ and $t = x$

• If $e$ and $11$ are coprime then, since $e$ divides both $3z+2t$ and $z - 3t$ , it follows that $e$ divides both $11z$ and $11t$ , so we deduce that $e$ divides both $z$ and $t$ . But that means that $e$ divides $1 = xt + yz$ . Since $e = z^2 + 2t^2 \ge 0$ , we deduce that $e=1$ , and hence $x \; = \; 3z + 2t \hspace{1.5cm} y \; = \; z - 3t \hspace{1.5cm} z^2 + 2t^2 \; = \; 1$ Thus we deduce that $t = 0$ and $z = \pm1$ , so that $x = 3z$ and $y = z$ .

Thus there are just four solutions in integers to these equations, namely $(x,y,z,t) = (1,0,3,1)\,,\,(-1,0,-3,-1)\,,\,(3,1,1,0)\,,\,(-3,-1,-1,0)$ . In all four cases we have $x^2+ y^2 + z^2 + t^2 = \boxed{11}$ .

For solving $\begin{cases} xz-2yt=3 \\ xt+yz=1 \end{cases}$ I'll do some algebraic manipulations involving complex numbers.

• Let $-2={(i \sqrt {2}) } ^ {2}$ , substituting on the first equation will lead to $xz+{(i \sqrt {2}) } ^ {2} yt=3$ .
• Multipling the second one by $(i \sqrt {2})$ , leading to $(i \sqrt {2})xt+(i \sqrt {2})yz=(i \sqrt {2})$ .
• Aligning the two equations $\begin{aligned} xz+{ (i\sqrt { 2 } ) }^{ { 2 } }yt=3 \\ (i\sqrt { 2 } )xt+(i\sqrt { 2 } )yz=i\sqrt { 2 } \end{aligned}$ and summing then up.
• Lefting us with $xz+(i\sqrt { 2 } )xt+{ (i\sqrt { 2 } ) }^{ { 2 } }yt+(i\sqrt { 2 } )yz=3+i\sqrt { 2 } \Rightarrow x(z+t\sqrt { 2 } i)+y\sqrt { 2 } i(z+t\sqrt { 2 } i)=3+\sqrt { 2 } i\Rightarrow (z+t\sqrt { 2 } i)(x+y\sqrt { 2 } i)=3+\sqrt { 2 } i$ .
• Applying moduli both sides wields $|(z+t\sqrt { 2 } i)(x+y\sqrt { 2 } i)|=|3+\sqrt { 2 } i| \Rightarrow |(z+t\sqrt { 2 } i)| \cdot |(x+y\sqrt { 2 } i)|=|3+\sqrt { 2 } i|$ .
• Using a complex number modulus, we have $\sqrt { (z ^ 2 +2 t ^2)\cdot (x^2+2y^2) } =\sqrt {11} \Rightarrow (z ^ 2 +2 t ^2)\cdot (x^2+2y^2) =11$
• Now for using some number theory, if $z, x, y, t \in \mathbb{Z}$ then their squares belong to $\mathbb{Z }$ too. So we have, in the last equation, a product of two integers resulting in a integer. There is only one product of integers resulting in $11$ which is $1 \cdot 11$ .
• $z^{ 2 }+2t^{ 2 }=1 \wedge x^{ 2 }+2y^{ 2 } = 11$ , and the symmetry of the equations will lead to the same answers, no need of breaking this to two or more cases.
• Testing for small integers, e.g. $0,1,2$ , equation $z^{ 2 }+2t^{ 2 }=1$ leads to $t=0, z= \pm 1$ and $x^{ 2 }+2y^{ 2 } = 11$ to $x=\pm 3, y=\pm 1$ .
• Thus $x^2+y^2+z^2+ t^2 \Rightarrow \boxed { 9+1+1+0=11}$

We work in $\mathbb{Z}[\sqrt{-2}]$ ; that is, numbers in the form $a + b \sqrt{-2}$ . If $p = a + b \sqrt{-2} \in \mathbb{Z}[\sqrt{-2}]$ , define $|p| = a^2 + 2b^2$ ; this is the square of the usual norm in complex field. We can prove that $|pq| = |p| \cdot |q|$ , and if $a, b$ are integers, then so as $|a + b\sqrt{-2}|$ .

Now, suppose we have $x + y \sqrt{-2}$ and $z + t \sqrt{-2}$ . Their product is $(xz - 2yt) + (xt + yz) \sqrt{-2}$ , or $3 + \sqrt{-2}$ . Thus we're trying to factorize this number.

Note that $|3 + \sqrt{-2}| = 3^2 + 2 \cdot 1^2 = 11$ . If $3 + \sqrt{-2}$ factors to $x + y \sqrt{-2}$ and $z + \sqrt{-2}$ , then 11 must also factor to $|x + y \sqrt{-2}|$ and $|z + t \sqrt{-2}|$ . Thus one of them is 11 and the other is 1. Without loss of generality, assume $|x + y \sqrt{-2}| = 11$ and $|z + t \sqrt{-2}| = 1$ .

Since $|z + t \sqrt{-2}| = z^2 + 2t^2$ , clearly we have $t = 0$ (any other integer $t$ makes the sum larger than 1) and $z^2 = 1$ . With a little casework, we also find that $x^2 = 9, y^2 = 1$ is the only integer solution to $|x + y \sqrt{-2}| = x^2 + 2y^2 = 11$ . Thus we have $x^2 + y^2 + z^2 + t^2 = 9 + 1 + 1 + 0 = \boxed{11}$ . In other words, this means $3 + \sqrt{-2}$ is irreducible, because if we try to factor it into a product of two numbers, one of the factors must be 1 or -1, so we don't simplify it at all. (Compare with $4 + \sqrt{-2} = (1 + \sqrt{-2})(2 - \sqrt{-2})$ .)

squaring both equations: $\begin{cases}x^2z^2-4xyzt+4y^2t^2&=9\\x^2t^2+2xyzt+y^2z^2&=1\end{cases}$

sum up (Eq1)² and ( $2\times$ Eq2²): $x^2(z^2+\mathbf 2t^2)+y^2(4t^2+\mathbf 2z^2)=9+\mathbf 2$ .

which can be factored as $(x^2+2y^2)(z^2+2t^2)=11$

Then $x^2,y^2,z^2,t^2$ are non negative integers, which yield a factorisation $11=1\cdot 11$ or $11=11\cdot 1$ . It is clear now that it suffices to check when $x^2,y^2,z^2,t^2$ have values in $\{0^2,1^2,2^2,3^2\}$ , thus the possible sums of form $a^2+2b^2$ :

$\begin{array}{cc|cccc} \text{sum} &a^2&0&1&4&9\\ b^2&2b^2 \\ \hline 0^2&0&0&\mathbf1&4&9\\ 1^2&2&2&3&6&\mathbf{11}\\ 2^2&8&8&9&-&-\\ \end{array}$

Therefore, $\{(x,y),(z,t)\}=\{(0,1), (1,3)\}$ and thus $x^2+y^2+z^2+t^2=0^2+1^2+1^2+3^2=11$ .

Just play around till you come up with the idea to put $t=0$ . Then it is easy but has many solutions. The smallest solution for which all are integers is $x=3,y=z=1$ which gives answer 11.