Post #3142159

\[(1+\epsilon)^2\leqslant q<\left(\frac{(2)\cdot(14)\cdot\left\{2k+1+(k-1)\epsilon\right\}}{9(k+2)\cdot(2+(1-k)\epsilon^{\prime})}\right)\leqslant \frac{5}{3}\] \[ \because(1+\epsilon)< \left(\frac{14}{9}\right)\approx q \leqslant \left(\frac{5}{3}\right)\] \[\textit{Verification by ChatGpt Let us verify the corrected inequality}\] \[\boxed{ n< \frac{28}{9\bigl(2+(1-k)\epsilon'\bigr)} \bigl(2k+1+(k-1)\epsilon\bigr). }\] carefully on actual abc triples. 🔷 Triple 1 $3+125=128$ Here:$k=4$,$n=7$, \epsilon\approx0.223, \qquad \epsilon'=0.25. Compute denominator: 9(2+(1-k)\epsilon')=9(2-3\cdot0.25) =9(1.25) =11.25. Bracket: 2k+1+(k-1)\epsilon = 9+3(0.223) = 9.669. Thus RHS: \frac{28}{11.25}(9.669) \approx2.489\times9.669 \approx24.07. Hence: \boxed{ 7<24.07 }true. 🔷 Triple 2 5+3\cdot13^2=2^9 Here:k=7, \qquad n=9, \epsilon\approx0.166, \qquad \epsilon'=0.125. Denominator: 9(2-7\cdot0.125) =9(1.125)=10.125. Bracket: 17+7(0.166) \approx18.162. RHS: \frac{28}{10.125}(18.162) \approx2.765\times18.162 \approx50.2. Thus: \boxed{ 9<50.2 }true.🔷 Eric Reyssat’s Triple 3 2+3^9\cdot109=23^5 Here: k=2, \qquad n=5, \qquad \epsilon\approx0.365, \qquad \epsilon'\approx0.148. Denominator: 9(2-0.148)=16.668. Bracket: 5+0.365=5.365. RHS: \frac{28}{16.668}(5.365) \approx1.679\times5.365 \approx9.01. Hence:\boxed{ 5<9.01 }true. 🔷 Now test a k=1 triple 1+2^5\cdot3\cdot5^2=7^4. Then: k=1, \qquad n=4. For k=1: (1-k)\epsilon'=0, \qquad (k-1)\epsilon=0. So RHS simplifies to \frac{28}{18}(3) = \frac{14}{9}\cdot3 = \frac{14}{3} \approx4.667. Thus: \boxed{ 4<4.667 } true.🔥 Final conclusion \textit{The expression is numerically correct for the tested abc triples, including both:} * k=1, * and k\ge2.\)