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Home » Solution: We seek integer solutions $ (x, y) $ to $ x^2 - y^2 = 2025 $. - Richter Guitar

Solution: We seek integer solutions $ (x, y) $ to $ x^2 - y^2 = 2025 $. - Richter Guitar

📅 May 4, 2026 👤 scraface

May 04, 2026

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📌 9\left(\frac{256}{81}n^2\right) + 16n^2 = \frac{256}{9}n^2 + 16n^2 = \frac{256 + 144}{9}n^2 = \frac{400}{9}n^2 = 25.

📌 Thus $ n^2 = \frac{25 \cdot 9}{400} = \frac{225}{400} = \frac{9}{16} \implies n = \pm \frac{3}{4} $.

📌 If $ n = \frac{3}{4} $, $ m = -\frac{16}{9} \cdot \frac{3}{4} = -\frac{4}{3} $. If $ n = -\frac{3}{4} $, $ m = \frac{4}{3} $.

📌 Check $ \vec{OG} \cdot (\vec{OA} + \vec{OB}) = 0 $: Both cases satisfy. However, the problem likely expects positive $ n $ (growth direction). Thus $ (m, n) = \left(-\frac{4}{3}, \frac{3}{4}\right) $.

📌 \boxed{\left( -\dfrac{4}{3}, \dfrac{3}{4} \right)}

📌 Question: The maximum imaginary part of a root of $ z^5 + z^3 + z + 1 = 0 $ can be expressed as $ \sin \theta $. Find $ \theta $ in degrees.

📌 Solution: Factor the polynomial:

📌 z^5 + z^3 + z + 1 = z^3(z^2 + 1) + 1(z + 1).

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