I am trying to find a geometric progression containing 2, 3 and 5 (the terms do not have to be consecutive). If there is no such progression, is it possible to prove this? Thanks in advance.
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1$$ar^n=2, ar^{n+x}=3, ar^{n+y}=5$$ – lab bhattacharjee Sep 03 '15 at 13:14
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Assume that $\{a_n\}_{n\in\mathbb{N}}$ is a geometric progression, $\frac{a_{n+1}}{a_n}=r$, and $$ a_m = 2,\qquad a_{m+i}=3,\qquad a_{m+j}=5. $$ That implies $\frac{5}{2}=r^j$ and $\frac{3}{2}=r^{i}$, hence: $$ \left(\frac{5}{2}\right)^i\cdot\left(\frac{2}{3}\right)^{j} = 1 $$ but that is equivalent to: $$ 2^{j-i} 5^i = 3^{j} $$ that contradicts the unique factorization theorem.
Jack D'Aurizio
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