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Home » æ¯è¾ææ°ï¼\(2x = 4\) - Aurero

æ¯è¾ææ°ï¼\(2x = 4\) - Aurero

📅 March 14, 2026 👤 scraface

Mar 14, 2026

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📌 Let \( k = 5a + 3 $. Substitute into the second:

📌 5a + 3 \equiv 4 \pmod{7} \implies 5a \equiv 1 \pmod{7}

📌 Multiply both sides by the modular inverse of 5 modulo 7. Since \( 5 \cdot 3 = 15 \equiv 1 \mod 7 \), the inverse is 3:

📌 a \equiv 3 \cdot 1 = 3 \pmod{7} \implies a = 7b + 3

📌 k = 5(7b + 3) + 3 = 35b + 15 + 3 = 35b + 18

📌 So \( k \equiv 18 \pmod{35} \)

📌 Now include the third congruence: \( k \equiv 2 \pmod{9} \)

📌 Let \( k = 35b + 18 \). Then:

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Solution: We are given two functional equations:
$ f(a + b + c) = f(a) + f(b) + f(c) $ for all real $ a, b, c $
$ f(ab) = f(a)f(b) $ for all real $ a, b $
We aim to find all such functions $ f : \mathbb{R} \to \mathbb{R} $.
Let’s try constant functions first. Suppose $ f(x) = c $ for all $ x $...
$ f(a + b + c) = c $, and $ f(a) + f(b) + f(c) = c + c + c = 3c $, so...
Also, $ f(ab) = 0 = 0 \cdot 0 = f(a)f(b) $, so the zero function works...
Now suppose $ f $ is not constant.
Let’s try additive functions satisfying both properties. From the addi...
f(a + b) = f(a) + f(b)

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