Peter wrote: Find all triples of positive integers $ (x, y, z)$ such that \[ (x + y)(1 + xy) = 2^{z}. \] Supose that $ x \leq y$,then $ x;y$ are odd;$ z \geq 2$ and exist $ m < z$ such that$ x + y = 2^m$ and $ 1 + xy = 2^{z - m}$ We have $ (1 + xy) - (x + y) = (x - 1)(y - 1) \geq 0$ so $ 2^{z - m} \geq 2^m$ or $ m \leq \frac {z}{2}$ *If $ x = 1$ then $ y = 2^m - 1;z = 2m$ *If $ x > 1$ then $ x \geq 3$=>$ 2^m = x + y \geq 6$ so $ m \geq 3$ We have $ x^2 - 1 = x(x + y) - (1 + xy) = 2^mx - 2^{z - m} = 2^m(x - 2^{z - 2m})$ So $ 2^m|x^2 - 1$ Note that $ (x - 1;x + 1) = 2$=>$ 2^{m - 1}|x - 1$ or $ 2^{m - 1}|x + 1$ From $ x \leq y$ we have $ 0 < x - 1 \leq \frac {x + y}{2} - 1 = 2^{m - 1} - 2 < 2^{m - 1}$ So $ 2^{m - 1}|x + 1$ From $ 1 < x \leq y$ then $ x + 1 < x + y = 2^m$ =>$ x + y = 2^{m - 1}$ or $ x = 2^{m - 1} - 1$ =>$ y = 2^{m - 1} + 1$=>$ 4z = 3m - 2$

Let we have $x+y=2^{a}=U,xy=2^{b}-1=V$ then $b\geq a$. With Viet,we have x,y are root of equation $t^{2}-Ut+V=0$ then we must have $k^{2}=2^{2(a-1)}-2^{b}+1$ or $(k-1)(k+1)=2^{b}(2^{2(a-1)-b}-1)$ note gcd(k-1,k+1)=2 hence $k=1$ or $k+1\geq 2^{b-1}\to k^{2}\geq 2^{2(b-1)}-2^{b}+1$ then $a\geq b$ All we have $a=b$ or $b+2=2a$ do next is easy

y=2^α+1 x=2^α−1

Let $x\ge y$. If $y=1$, then $x=2^m-1, z=2m$ give solution $(2^m-1,1,2m)$. $y>1\to x=2^{m-1}+t,y=2^{m-1}-t, x+y=2^m, m\ge 3$ and $2^{2m-2}-t^2=2^{z-m}-1$. Obviosly $t=1\to z=3m-2$ and $(x,y,z)=(2^{m-1}+1,2^{m-1}-1,z=3m-2), m\ge3$ is new solution with $x>y$. $1<t<2^m-1$ is odd and $t^2-1=2^{z-m}(2^{3m-2-z}-1)$. Let $s=\frac{t-1}{2}$, then $1\le s<2^{m-1}-1$ and $s(s+1)=2^{z-m-2}(2^{3m-2-z}-1)$. If $s$ even, then $s=2^{z-m-2}a, s+1=b, ab=2^{3m-2-z}-1$. Let $z=m+2+k, k\ge 2$, then, $b=2^ka+1$ and $2^{2m-4-k}-1=a(2^ka+1)$. If $a=1$, then $k=1,s=2\to t=5,2m-5=2$ - contradition. If $a=1\mod 4$ then $2^{2m-4-k}-1=3\mod 4\to k=1\to a(2a+1)=2^l-1, l=2m-5\to (4a+1)^2+7=(2^{m-1})^2$ have not solution. If $a=3\mod 4$, then $b=1\mod 4$, therefore $k\ge 2$ and $a(2^ka+1)+1=2^{2m-4-k}\to 2^{k+2}-1=2^{2m-2}-(2^{k+1}a+1)^2>2^{k+2}a+2$ have not solution. If $s$ odd, then $s+1=2^ka, s=b=2^ka-1$ and $2^{2m-4-k}-1=a(2^ka-1)$ or $(2^{k+1}a-1)^2+(2^{k+2}-1)=(2^{m-1})^2$. Because difference $y^2-x^2\ge 2x+1$ must be $2^{k+2}-1\ge (2^{k+2}a-1)\to a=1, b=2^k-1\to k=m-2,z=2m, t^2-1=2^m(2^{m-2}-1)\to t=2^{m-1}-1,x=2^m-1,y=1$ - old solution. Therefore all solutions are $(x,y,z)=(1,2^m-1,2m),(2^m-1,1,2m),$ and $ (2^{m-1}+1,2^{m-1}-1,3m-2),(2^{m-1}-1,2^{m-1}+1,3m-2), m\ge 3.$

equation implies that $x+y=2^{z_0},xy+1=2^{z_1}$ WLOG assume that $x\le y$ now if $x=1 \implies (1+y)=2^{\frac{z}{2}}$ then one possible solution is $(1,2^{z}-1,2z)$ so now because $x,y$ should be odd then $x\ge 3 \implies x+y \le xy+1 \implies x+y \mid xy+1 \implies$ $x+y=2^{z_0} \mid x^2-1 =(x+1)(x-1)$, it's easy to see $(x-1,x+1)=2$ now we have to check two case: I) $x+y \mid 2(x+1)$ now as $2(x+y) > 2(x+1)$ so $x+y=2(x+1) \implies y=x+2$ $\implies 2x+2=2^{z_0},xy+1=(x+1)^2=2^{z_1}$ so one other possible solution is $(2^{z}-1,2^{z}+1,3z+1)$ II) $x+y \mid 2(x-1) $ now as $2(x+y) > 2(x-1)$ implies $x+y=2x-2 \implies y=x-2$ which is a contradiction with $y \ge x$ so no solution in this case so the solutions of the equation are: $(2^{z}-1,2^{z}+1,3z+1)$ and $(1,2^{z}-1,2z)$