Tag Info

• 341
Accepted

Quadratic Diophantine Equation $x^2 + 2y^2 = 2013$

The equation is not solvable modulo $8$, hence there are no integer solutions. You can see this by considering $x^2\equiv 0,1$ or $4$ modulo $8$ for every $x$.
• 78.4k
Accepted

Solving the nonlinear Diophantine equation $x^2-3x=2y^2$

Multiply through by $4$ and complete the square, $(2x-3)^2 - 8 y^2 = 9,$ so that $(2x-3)^2 + y^2 \equiv 0 \pmod 3.$ It follows that both $(2x-3)$ and $y$ are divisible by $3,$ therefore $x$ as well. ...
• 130k

First 30 solutions of Pell's equation.

Floating point precision is not sufficient for the larger solutions. You should probably look at using a recursive definition for the $k_n$ and $z_n$. They should satisfy $a_{n+2}=18a_{n+1}-a_{n}$ (...
• 4,757
Accepted

Can $n+1$ , $2n+1$ , $3n+1$ all be perfect squares , if $n$ is a positive integer?

If this happens, then $1, n+1, 2n+1, 3n+1$ is a four-term arithmetic progression of perfect squares. But it can be shown (painfully, using elliptic curves, as demonstrated at this link - see Theorem ...
• 113k
Accepted

Prove $(1+\sqrt{2})^k= \sqrt{n}+\sqrt{n-1}$

This follows from $(1-\sqrt{2})^k\,(1+\sqrt{2})^k=(-1)^k$ for every integer $k$. So, if $a_k$ and $b_k$ are integers such that $(1+\sqrt{2})^k=a_k+b_k\sqrt{2}$, then $a_k-b_k\sqrt{2}=(1-\sqrt{2})^k$, ...
• 48.4k
Accepted

• 2,833

• 47.9k

Integer solutions of $y^{2} = 5x^{2} + 17$

Your attempt cannot work, since reducing modulo an integer (here $5$) irreversibly wipes out the distinction between positive an negative numbers; one can never apply arguments based on comparison "...
• 107k

Find all primes $p$ and $q$ such that $p^2-2q^2=1.$

From your last expression we have $$\tag1(p+1)(p-1)=2\cdot[\;2(2d^2+2d)+1\;].$$ If $p=2$ then you get from the original expression that $2q^2=3,$ which is impossible. Therefore $p>2$ and hence $p$ ...
• 3,309

Find all primes $p$ and $q$ such that $p^2-2q^2=1.$

Since $p^2-2q^2=1$, $p$ must be odd. Furthermore, $q$ must be even; if $q$ were odd, then $p^2-2q^2$ would be $3$ mod $4$. Thus, $q=2$ and then $p=3$.
• 326k

General solution of Pell's equation

It's well know that the solution of the Pell's equation satisfy the reccurence relation: $$x_{k+1} = x_1x_k + y_1ny_k$$ $$y_{k+1} = y_1x_k + x_1y_k$$ This was found by Brahmagupta and it holds ...
• 34.9k

How to show that $x^2 - 37y^2 =2$ does not have integer solutions

There are no integer solutions to $$x^2 - 37 y^2 \equiv 2 \pmod 4$$ as $$x^2 - 37 y^2 \equiv x^2 - y^2 \pmod 4$$
• 130k
there are infnitely many postive integer $n$ such $\lfloor \sqrt{7}\cdot n \rfloor=k^2+1(k\in \Bbb{Z})$
I get that there are an infinite number of $n$ such that $\lfloor n\sqrt{d} \rfloor =k^2-1$, not $k^2+1$. However, for $d$ such that there are solutions to $x^2-dy^2 = -3$, such as $d=7$, then there ...