Exercise 7.8.13 ($x^2 - 34 y^2 = -1$ is not solvable, but has solutions modulo $m$ for every $m$)

Proof. We decompose $34y^2+1$ and $34y^2-1$ in prime numbers for $y=1,2,3,4,5,6$:

The first square is $34\cdot 6^2+1=35^2$. Therefore the first positive solution of $x^2-34y^2=\pm 1$ is $(35,6)$, which is a solution of $x^2-34y^2=1$.

By Problem 1, $x^2-34y^2=-1$ is not solvable, otherwise there would exist a positive solution of $x^2-34y^2=-1$ for some $y\in [[1,5]]$.

(Alternatively, we can use with more efficiency the Python method “pell_fermat” of Problem 9, which give the least positive solution of $x^2-34y^2=\pm 1$: we obtain $(35,6)$. This shows that $x^2-34y^2=-1$ is not solvable.)

Since

The pairs $(5∕3,1∕3)$ and $(3∕5,1∕5)$ are rational solutions of $x^2-34y^2=-1$.

Let $m$ be a positive integer. Using the first of these rational solutions, $5^2-34\cdot 1^2\equiv -3^2(\mathit{mod}m)$. Suppose that $3\nmid m$. Then $3$ has in inverse modulo $m$. If we write $ā$ the class of an integer $a$ in $\mathbb{Z}∕\mathit{m\mathbb{Z}}$, we obtain

Put $\overline{x_1}=\bar{5}\cdot {\bar{3}}^{-1},\overline{y_0}={\bar{3}}^{-1}$, where $x_1,y_1$ are integers. Then

The congruence $x^2-34y^2\equiv -1(\mathit{mod}m)$ has a solution provided that $3\nmid m$.

Similarly, if $5\nmid m$, then $5$ has an inverse modulo $m$. From ${\bar{3}}^2-\overline{34}\cdot {\bar{1}}^2=-{\bar{5}}^2$, we obtain that $(\overline{x_2},\overline{y_2})=(\bar{3}\cdot {\bar{5}}^{-1},,{\bar{5}}^{-1})$ gives a solution of $x^2-34y^2\equiv -1(\mathit{mod}m).$

Now suppose that $m$ is any integer, where $m>1$. Write $m=3^{\alpha }{5}^{\beta }u$, where $3\nmid u,5\nmid u$. Put $m_1=5^{\alpha }$ and $m_2=3^{\beta }u$. Then $m=m_1{m}_2$, where $m_1\wedge m_2=1$.

Since $3\nmid m_1$, there are some integers $x_1,y_1$ such that $x_1^2-34y_1^2\equiv -1(\mathit{mod}{m}_1)$, and since $5\nmid m_2$, there are some integers $x_2,y_2$ such that $x_2^2-34y_2^2\equiv -1(\mathit{mod}{m}_2)$.

By the Chinese Remainder Theorem, since $m_1\wedge m_2=1$, there exist some integers $x$ and $y$ such that

Then $x^2-34y^2\equiv -1(\mathit{mod}{m}_1)$ and $x^2-34y^2\equiv -1(\mathit{mod}{m}_2)$, where $m_1\wedge m_2=1$. Hence $x^2-34y^2\equiv -1(\mathit{mod}m)$.

In conclusion, the equation $x^2-34y^2=-1$ has no solution in $\mathbb{Z}$, but has solutions in $\mathbb{Q}$ and in $\mathbb{Z}∕\mathit{m\mathbb{Z}}$ for every $m>1$. □