prove if n - natural number divide number $34x^2-42xy+13y^2$ then n is sum of two square number

Question

prove if n - natural number divide number $34x^2-42xy+13y^2$ where x,y are relatively prime then n is sum of two square number.

I don't know what is going on in this exercise. I will be grateful for explanation.

Answer 1

A start: The natural number $n$ divides $34(34x^2-42y+13y^2)$, which is equal to $$(34x-21y)^2+y^2.$$ Now use what you know about the prime factorization of numbers that are the sum of two squares.

Added: We show that if $x$ and $y$ are relatively prime, then any prime divisor of $(34x-21y)^2+y^2$ is either $2$ or of the form $4k+1$.

Since $2$ is a sum of two squares, and any prime of the form $4k+1$ is a sum of two squares, we get that $n$ is a product of sums of two squares. Thus by the Brahmagupta-Fibonacci-Diophantus identity, $n$ is a sum of two squares.

So let us look at the prime divisors of $(34x-21y)^2+y^2$. Let $d$ be the gcd of $34x-21y$ and $y$. Then $d$ divides $34x$, and therefore since $x$ and $y$ are relatively prime, it follows that $d$ divides $34$.

Thus $34x-21y=dA$ and $y=dB$, where $A$ and $B$ are relatively prime. It follows that any prime divisor of $(34x-21y)^2+y^2$ is $2$ or $17$, or a prime divisor $p$ of $A^2+B^2$. Since $A$ and $B$ are relatively prime, it follows by standard theory that $p=2$ or $p$ is of the form $4k+1$. This completes the proof.

Answer 2

They are asking about Gauss reduction. The given form is positive and of discriminant $−4.$ That means there is an invertible (determinant 1) change of variables in $SL_2 \mathbb Z,$ call it $$ P = \left( \begin{array}{rr} \alpha & \beta \\ \gamma & \delta \end{array} \right) $$ that takes one form to the other, and preserves gcd. If $u=2x+3y, v=3x+5y,$ then $$34u^2−42uv+13v^2=x^2+y^2$$ Backwards, $$(5x−3y)^2+(−3x+2y)^2=34x^2−42xy+13y^2.$$
Here we are looking at $$ H = \left( \begin{array}{rr} 34 & -21 \\ -21 & 13 \end{array} \right) $$ and $$ P = \left( \begin{array}{rr} 2 & 3 \\ 3 & 5 \end{array} \right) $$ Taking $$ Q = P^{-1} = \left( \begin{array}{rr} 5 & -3 \\ -3 & 2 \end{array} \right), $$

The first displayed line equation $u=2x+3y, v=3x+5y,$ $34u^2−42uv+13v^2=x^2+y^2$ is $$ P^T H P = I, $$ the second is $$ Q^T Q = H. $$ The second ought to make you think of Cholesky, only with integers required.

It happens that $P$ and so $Q$ are symmetric so the transpose signs are redundant, not important.

EDIT: Gauss reduction is a step by step procedure intimately linked with the Euclidean algorithm. At each step, we are changing the symmetric matrix in the middle by multiplying on the right by either some $$ P_j = \left( \begin{array}{rr} 1 & \beta \\ 0 & 1 \end{array} \right) $$ or $$ P_j = \left( \begin{array}{rr} 0 & -1 \\ 1 & 0 \end{array} \right) $$ and keeping the running product $P_1 P_2 \cdots P_r = P.$ The changes on the triple of coefficients go this way: $$ \langle 34, -42, 13 \rangle, $$ $$ \langle 13, 42, 34 \rangle, $$ $$ \langle 13, -10, 2 \rangle, $$ $$ \langle 2, 10, 13 \rangle, $$ $$ \langle 2, 2, 1 \rangle, $$ $$ \langle 1, -2, 2 \rangle, $$ $$ \langle 1, 0, 1 \rangle. $$ The particular string $P_1 P_2 \cdots P_6 = P$ that accomplishes this is $$ \left( \begin{array}{rr} 0 & -1 \\ 1 & 0 \end{array} \right) \left( \begin{array}{rr} 1 & -2 \\ 0 & 1 \end{array} \right) \left( \begin{array}{rr} 0 & -1 \\ 1 & 0 \end{array} \right) \left( \begin{array}{rr} 1 & -2 \\ 0 & 1 \end{array} \right) \left( \begin{array}{rr} 0 & -1 \\ 1 & 0 \end{array} \right) \left( \begin{array}{rr} 1 & 1 \\ 0 & 1 \end{array} \right) = \left( \begin{array}{rr} 2 & 3 \\ 3 & 5 \end{array} \right) $$