166 views

$p^2+1=q^2+r^2$. Strange phenomenon of primes

Problem: Find prime solutions to the equation $p^2+1=q^2+r^2$ I welcome you to post your own solutions as well I have found a strange solution which I can't understand why it works(or what's the ...
188 views

Positive even numbered integer solutions of $y=n^2-m^2-x^2$

Prove that no integer $x$ exists where $y=n^2-m^2-x^2$ has solutions: For all even integer values of $y$ in the range $2\le y \le 2x+1$ where $x$ is odd. For all odd integer values of $y$ in the ...
85 views

Number of integer points on a rotational hyperboloid of two sheets.

There are many integer points on the hyperboloid of two sheets $x^2+y^2-z^2=-1$. (0,0,1), (2,2,3), (4,8,9),... Let us denote such set as H. I will consider only the upper sheet $z>0$, but ...
141 views

$x^2+y^2=z(4z+1)$ solutions

For a small project I am working on, I wish to find the solutions for $$x^2+y^2=z(4z+1)$$ in natural numbers $x,y,z$. I wish to automate finding solutions for $z$ up to a maximum value as efficient as ...
1k views

Showing that $m^2-n^2+1$ is a square

Prove that if $m,n$ are odd integers such that $m^2-n^2+1$ divides $n^2-1$ then $m^2-n^2+1$ is a square number. I know that a solution can be obtained from Vieta jumping, but it seems very different ...
6k views

Solution of Diophantine equation

Find all integral solutions of $x^2+1= y^2+z^2$. Actually I have to find all integral solution of $a(a+1)=b(b+1)+c(c+1)$. I reduced this in the above form I.e., $(2a+1)^2+1= (2b+1)^2+(2c+1)^2$ .
159 views

Solving $x^2 + y^2 = 1 + z^4$ with (x,y,z) = 1 and z < x < y

I have a computer programming problem where I need to find n many sets of integers that meet the condition $x^2 + y^2 = 1 + z^4$ with (x,y,z) = 1 and z < x < y I can do this relatively easily ...
The diophantine equation $z^2=a^2+bx^2+cy^2$
Is there a way to obtain (enumerate) the integer solutions $(x,y,z)$ of the following quadratic Diophantine equation $z^2=a^2+bx^2+cy^2$ where $a$ is an integer and $b, c$ are positive integers? I ...