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  https://github.com/MyJestor/An-Alternative-to-Chinese-Remainder-Theorem- # An Iterative Alternative to Classical Chinese Remainder Theorem ## Background The classical CRT requires pairwise coprime moduli. I've developed an iterative algorithm that generalizes this. ## Main Result The algorithm solves any system of linear congruences by: 1. Combining congruences pair-wise iteratively 2. Using GCD and LCM properties 3. Requiring only that solutions exist (no coprimality condition) ## C...

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  https://github.com/MyJestor/An-Alternative-to-Chinese-Remainder-Theorem-

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  In continuation with above toot… ### 2. https://doi.org/10.1080/07468342.2002.11921953 This DOI belongs to *The College Mathematics Journal* (2002). Despite multiple searches, the exact title and full text could not be retrieved from open sources (the article is behind a paywall / not freely indexed in the results returned). From the journal and year, it is almost certainly a classroom note or short article related to congruences or the Chinese Remainder Theorem. The standard “method of succe...

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  ChatGPT an AI, says:I examined both DOIs. ### 1. https://doi.org/10.1080/00029890.1952.11988142 **Title:** *The General Chinese Remainder Theorem* **Author:** Øystein Ore **Journal:** *The American Mathematical Monthly*, Vol. 59, No. 6 (June–July 1952), pp. 365–370. Ore gives a generalization of the classical Chinese Remainder Theorem that works **even when the moduli are not pairwise coprime**. He states necessary and sufficient conditions for a solution to exist and provides a general for...

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  \[Let \quad a^n+b^n=c^n \quad be \quad true \quad for\quad \forall n\geqslant 3\] \[\textit{then there exists a positive integer k such that}\quad c<c^2<c^3<....<c^k<a^n <b^n\] \[and \quad c<c^2<c^3<....<c^k<a^n <b^n<c^{k+1}\] \[If \quad a^n+b^n=c^n \quad is \quad true \quad \forall n\geqslant 3 \quad then\] \[a^n=cq_1+r_1\quad \textit{by Euclid's Division Lemma}\quad 0\leqslant r_1<c\] \[...

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  https://math.stackexchange.com/questions/733754/visually-stunning-math-concepts-which-are-easy-to-explain

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  You wanted the proof of \[(m+q)(n+q)=2mn+q^2\] \[(m+q)(n+q)<2mn+\frac{mn}{k+2}\] \[(m+q)(n+q)<mn(2+\frac{1}{3})\] \[\left(\frac{m+q}{m}\right)\cdot \left(\frac{n+q}{n}\right)<\frac{7}{3}\] \[2\left(\frac{m+q}{m}\right)\cdot \left(\frac{n+q}{n}\right)<\frac{14}{3}\] \[\left\{\left(\frac{m+q}{m}\right)+ \left(\frac{n+q}{n}\right)\right\}^2=3^2\] \[\left(\frac{m+q}{m}\right)^2+ 2\left(\frac{m+q}{m}\right)\cdot \left(\frac{n+q}{n}\right)+\left(\frac{n+q}{n}\right)^2=3^2\...

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  \[(𝑘+2)𝑘ε^′+(𝑘+1)ε=1 \] DOI: https://doi.org/10.17605/OSF.IO/KR2MQ

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  \subsection{Numerical Verification of the Expression} The following table shows the numerical verification of the expression \[ \frac{5}{3} \approx \frac{m+q}{m} \approx \frac{6 + (k^2 + 1)\epsilon' + k\epsilon}{4} \] on several abc triples with different values of \(k\). \begin{table}[h] \centering \begin{tabular}{|c|l|c|c|c|c|} \hline \(k\) & Triple (example) & Actual \(\frac{m+q}{m}\) & Right side \(\frac{6 + (k^2+1)\epsilon' + k\epsilon}{4}\) &am...

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  If you want to know truth, learn mathematics.

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  \[(1+\epsilon)^2\leqslant q<\left(\frac{(2)\cdot(14)\cdot\left\{2k+1+(k-1)\epsilon\right\}}{9(k+2)\cdot(2+(1-k)\epsilon^{\prime})}\right)\leqslant \left(\frac{5}{3}\right)\] \[\qquad \because(1+\epsilon)< \left(\frac{14}{9}\right)\approx q \leqslant \left(\frac{5}{3}\right)\]

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  \[(1+\epsilon)^2\leqslant q<\left(\frac{(2)\cdot(14)\cdot\left\{2k+1+(k-1)\epsilon\right\}}{9(k+2)\cdot(2+(1-k)\epsilon^{\prime})}\right)\leqslant \frac{5}{3}\] \[ \because(1+\epsilon)< \left(\frac{14}{9}\right)\approx q \leqslant \left(\frac{5}{3}\right)\] \[\textit{Verification by ChatGpt Let us verify the corrected inequality}\] \[\boxed{ n< \frac{28}{9\bigl(2+(1-k)\epsilon'\bigr)} \bigl(2k+1+(k-1)\epsilon\bigr). }\] carefully on actual abc triples. 🔷 Triple 1 $...