2 votes

$x^5=y^2+10$ has no solutions

Josef Blass proved in Math. Comp. 30 (1976) 638-640 the following $\textbf{Theorem.}$ If $k$ is a square-free positive integer for which $k \not\equiv 7 \pmod 8 $ and $(h(-k),5)=1$, then the ...
Tomita's user avatar
  • 2,366
2 votes

$x^5=y^2+10$ has no solutions

A PARTIAL ANSWER. $x,y$ should have the same parity; if $x$ is even then $x^5=y^2+10\implies 0\equiv y^2+10\pmod{16}$ which is not possible because $y$ is even and its square can be only $0$ or $4$ ...
Piquito's user avatar
  • 29.9k
2 votes

How to prove that the Diophantine equation $(x+y)(x+y+2)=10xy$ has no positive integer solutions

The resulting Diophantine equation is $$y^2 + y(2 - 8x) + x^2 + 2x = 0 \tag{1}\label{eq1B}$$ As indicated in Bill Dubuque's comment, we can use Standard Vieta jumping to prove there are no positive ...
John Omielan's user avatar
  • 48.3k
2 votes

Diophantine equation $2x^4 + 2x^2 y^2 + y^4 = z^2$

$2x^4+2x^2y^2+y^4=z^2 \text{ for integers } x, y, z$ Quite long solution, and lots to edit… Case 1. $x=0$: $y^4=z^2, (x, y, z) = (0, t, \pm t^2)$ is a solution. Case 2. $x\ne0, y=0:$ $2x^4=z^2$, No ...
RDK's user avatar
  • 2,763
2 votes

Diophantine equation $2x^4 + 2x^2 y^2 + y^4 = z^2$

The only integer solutions of $$2x^4 + 2x^2 y^2 + y^4 = z^2\tag1$$ are $(x,y,z)=(0,s,\pm s^2)$ where $s$ is any integer. Proof : Let us first consider the case $xyz=0$. If $x=0$, then $z=\pm y^2$. ...
mathlove's user avatar
  • 141k
2 votes

Diophantine equation $2x^4 + 2x^2 y^2 + y^4 = z^2$

The question is Find all integer solutions $(x,y,z)$ to the equation $$ 2x^4 + 2x^2 y^2 + y^4 = z^2 $$ Divide both sides by $y^4$ to get $$ 2r^4 + 2r^2 + 1 = s^2 $$ where $\,r=x/y\,$ and $\,s=z/y^2.\...
Somos's user avatar
  • 35.3k

Only top scored, non community-wiki answers of a minimum length are eligible