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mukushla

  • 3 years ago

Find all triples \((x,m,n)\) of positive integers satisfying the equation \[x^m=2^{2n+1}+2^n+1\]

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  1. mukushla
    • 3 years ago
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    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 \( (23,2,4)\)

  2. mukushla
    • 3 years ago
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