Đặt \(2^k+2^4+2^7=q^2\left(q\in\text{ℕ},q>0\right)\)
\(\Leftrightarrow2^k+12^2=q^2\)
\(\Leftrightarrow\left(q-12\right)\left(q+12\right)=2^k\)
Vì q∈ℕ* nên q+12>q-12.
Đặt \(q+12=2^n,q-12=2^m\left(n,m\in\text{ℕ*},n>m\right)\);n+m=k.
\(\Rightarrow2^n-2^m=12-\left(-12\right)=24\)
\(\Leftrightarrow2^m\left(2^{n-m}-1\right)=24\)
Có: \(2^{n-m}-1\)lẻ
\(\Rightarrow2^m\left(2^{n-m}-1\right)=24=8.3\)
\(\Rightarrow2^m=8\Rightarrow m=3\left(TM\right)\)
\(\Rightarrow n=5\Rightarrow k=8\)
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