## mukushla Group Title Find all triples $$(x,m,n)$$ of positive integers satisfying the equation $x^m=2^{2n+1}+2^n+1$ 2 years ago 2 years ago

Well note that $$x$$ is an odd number. $$m=1$$ is an obvious answer so i want to work on $$m>1$$.............. firstly suppose that $$m$$ is an odd number $$m=2k+1$$ where $$k \ge 1$$ we have $2^{n}(2^{n+1}+1)=x^{m}-1=(x-1)(x^{m-1}+...+1)$$$(x^{m-1}+x^{m-2}+...+x+1)=(x^{2k}+x^{2k-1}...+x+1)$$ is an odd number since $$x$$ is odd so $$x-1$$ divides $$2^n$$ then $$x-1 \ge 2^n$$ and $$x\ge 1+2^n$$ hence : $2^{2n+1}+2^{n}+1=x^{m}\geq (2^{n}+1)^{3}=2^{3n}+3\times 2^{2n}+3\times 2^{n}+1$and this is a contradiction.....so $$m$$ is even now let $$m=2k$$ then $$2^{n}(2^{n+1}+1)=x^{2k}-1=(x^k-1)(x^k+1)=(y-1)(y+1)$$ and $$y$$ is odd. $$y-1$$ and $$y+1$$ are even factors so $$\gcd(y-1,y+1)=2$$ exactly one of them divisible by $$4$$. Hence $$n ≥ 3$$ and one of these factors is divisible by $$2^{n−1}$$ but not by $$2^{n}$$. So we can write $$y = 2^{n−1}s +t$$ , $$s$$ odd , $$t= ±1$$. Plugging this into the original equation we obtain $$2^n(2^{n+1}+1)=(2^{n−1}s +t)^2-1=2^{2n−2}s^2+2^n st$$ or, equivalently $$2^{n+1}+1=2^{n−2}s^2+ st$$ Therefore $$1-st=2^{n−2}(s^2-8)$$ For $$t= 1$$ this yields $$s^2 − 8 ≤ 0$$, so $$s = 1$$, which fails to satisfy original equation. For $$t=-1$$ equation gives $$1+s=2^{n−2}(s^2-8) \ge 2(s^2-8)$$ so $$2s^2-s-17 \le 0$$ and $$s \le 3$$ we know that $$s$$ is odd so $$s=3$$. put this in the last equation with $$t=-1$$ gives $$n=4$$. back to the original equation with $$n=4$$ u have $x^m=529=23^2$ so $$x=23$$ and $$m=2$$ the only solutions are $$(x,m,n)=(2^{2l+1}+2^l+1, 1,l)$$ and $$(23,2,4)$$