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📅 February 27, 2026 👤 scraface

Feb 27, 2026

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$ xu - yv = 13 $,
$ xv + yu = -2 $.
From $ u = 2 - x $, $ v = 4 - y $. Substitute:
$ x(2 - x) - y(4 - y) = 13 \Rightarrow 2x - x^2 - 4y + y^2 = 13 $,
$ x(4 - y) + y(2 - x) = 4x - xy + 2y - xy = 4x + 2y - 2xy = -2 $.
First equation: $ -x^2 + y^2 + 2x - 4y = 13 $.
Second equation: $ 4x + 2y - 2xy = -2 \Rightarrow 2x + y - xy = -1 $.
Let $ y = -1 - 2x + xy $ â circular. Solve second: $ 4x + 2y = 2xy -...
Let $ y = x - 1 - rac{1}{x} $ â not helpful. Try numerical values....
$ a^2 - b^2 = -64 $,

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\sum_{k=1}^{24} 3^k = 3 \cdot \frac{3^{24} - 1}{3 - 1} = \frac{3}{2} (...
But working modulo $7$, we can compute the sum directly using cyclicit...
3^1 &\equiv 3 \mod 7 \\
3^2 &\equiv 9 \equiv 2 \mod 7 \\
3^3 &\equiv 3 \cdot 2 = 6 \mod 7 \\
3^4 &\equiv 3 \cdot 6 = 18 \equiv 4 \mod 7 \\
3^5 &\equiv 3 \cdot 4 = 12 \equiv 5 \mod 7 \\
3^6 &\equiv 3 \cdot 5 = 15 \equiv 1 \mod 7 \\
So the powers of $3 \mod 7$ repeat every 6 terms. The cycle length is...
The sum of one full cycle:

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