MHB No Positive Integer Solution for $4xy - x - y = z^2$

[kaliprasad]

show that the equation $4xy - x - y = z^2$ has no positive integer solution

 

[individ]

Spoiler

I decided in the General form:
It's pretty old equations that are solved by Euler.
the equation:

$$aXY+X+Y=Z^2$$

If we use the solutions of Pell's equation: $$p^2-acs^2=\pm1$$

Solutions can be written:

$$X=\pm(c+1)s^2$$

$$Y=\pm(c+1)cs^2$$

$$Z=ps(c+1)$$

$$c$$ - We ask ourselves. While the formula and can be written differently.

Equation: $$p^2-4cs^2=-1$$ has no solutions.
Because : $$\frac{p^2+1}{4}$$ may not be an integer.

 

[kaliprasad]

Spoiler

I decided in the General form:
It's pretty old equations that are solved by Euler.
the equation:

$$aXY+X+Y=Z^2$$

If we use the solutions of Pell's equation: $$p^2-acs^2=\pm1$$

Solutions can be written:

$$X=\pm(c+1)s^2$$

$$Y=\pm(c+1)cs^2$$

$$Z=ps(c+1)$$

$$c$$ - We ask ourselves. While the formula and can be written differently.

Equation: $$p^2-4cs^2=-1$$ has no solutions.
Because : $$\frac{p^2+1}{4}$$ may not be an integer.

Because it is a challenge question complete solution is required

 

[individ]

And what answer do not like? The formula is. What problems?

 

[Albert1]

show that the equation $4xy - x - y = z^2$ has no positive integer solution

Spoiler

if $xy=0$ then the solutions will be :
$(x,y,z)$=(0,$-k^2$,$\pm k)$
or$ (x,y,z)$=($-k^2$,0,$\pm k)$
(here $k \in Z$) has no positive solution
if $xy\neq 0$ then $4xy-x-y$ can not be prime
we set $xy=p---(1)$
and $x+y=q--(2)$ then
$4p-q=2(2p-\dfrac {q}{2})=z^2$
$\therefore z=2 $ and $2p-\dfrac{q}{2}=2$
or $q=4m,p=m+1\,(m\in N)$
put $q=4m,$ and $p=m+1$ to (1)and (2)we get:
$y=2m\pm \sqrt {4m^2-m-1}$ $\notin N$
and the proof is done

 

[individ]

Not the correct solution.
The condition for which solutions need to come from the solution, not the solution of the conditions that we want to impose.
Very often many so decide.
Your mistake is easily detected if you need to find a solution to the equation:

$$aXY-X-Y=Z^2$$

How do You then the solution of this equation can find?

 

[kaliprasad]

$4xy-x - y = z^2$

Spoiler

multiply by 4 and add 1 on both sides

$16xy-4x - 4y + 1 = 4z^2+ 1$

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

let p be a prime divisor of 4x- 1

$(2z)^2 \equiv - 1 \, mod\, p$

as per Fermats Little theorem

$(2z)^{p-1} \equiv 1 \, mod\, p$

hence
$(2z)^{p-1} \equiv (2z^2)^{\frac{p-1}{2}}\, mod\, p\equiv (-1)^{\frac{p-1}{2}} \equiv 1\, mod\, p $

hence $ p \equiv 1 \,mod\,4$
hence all factors of 4x -1 are of the form 4y+1 which is a contradiction

so no solution

 

[individ]

Leave the rest of modular arithmetic.
Should equation to solve.
You generally solutions not, and they are when the number is negative.
I asked about the decision not such equation, and such?

$$aXY-X-Y=Z^2$$

 

[kaliprasad]

Leave the rest of modular arithmetic.
Should equation to solve.
You generally solutions not, and they are when the number is negative.
I asked about the decision not such equation, and such?

$$aXY-X-Y=Z^2$$

The above is partial information and does not contain any details and hence it is irreverent