Solve in integers.

Question

Solve in integers:

$3x - 3y = xy - 5$
$y^2 = 5x + 6$
$1+p+p^2+p^3=3^n$

For second, i think that infinite number of solutions.

Maybe $x = 5$ $n^2 - 8 n + 2$, $y = 4 - 5 n$, n element $\mathbb{Z}$

Answer 1

$(1.)$

$$3x - 3y = xy-5$$ $$3x - xy -3y = -5$$ $$x(3-y)+3(3-y) = 4$$ $$(x+3)(3-y) = 4 $$ $$(x,y) = (-2,-1) \,\, ,(-1,1)\,\, ,(1,2) \,\, , (-4,7) \,\, ,(-5,5) \,\, , (7,4)$$

$(2.)$

This equation has infinitely many solutions over integers.

$(3.)$

First we check that the trivial case $ p = n = 0$ which satisfies the equation. $$1+p+p^2+p^3 = 3^n$$ $$(p^2 + 1)(p+1) = 3^n$$

Let $(p^2 + 1) = 3 ^q$ and $(p + 1) = 3^r$ Then ,

$$p^2 + 1+ p+1 = 3^q + 3^r$$ $$ p(p+1) + 2 = 3^q + 3^r$$

Now notice that R.H.S is divisible by $3$ , implying that L.H.S is also divisible by $3.$ This is only possible if $ p^2 + p \equiv 1\mod 3.$ But it is easy to observe that there is no value of $p$ satisfying this relation . Hence the only possible solution is $(0,0).$

Answer 2

$$x=\dfrac{5-3y}{y-3}=\dfrac{5-3(y-3)+9}{y-3}=-3+\dfrac{14}{y-3}$$

So, $y-3$ must divide $14$

As $(y,5)=1$

If $y=5z\pm1$

$$5x+6=25z^2\pm10z+1\iff x=5z^2\pm2z-1$$

What if $y=5z\pm2$