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Home » There is at least one multiple of 3 → $ 3 \mid $ product - Esdistancia

There is at least one multiple of 3 → $ 3 \mid $ product - Esdistancia

📅 March 17, 2026 👤 scraface

Mar 17, 2026

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📌 This is the **Cauchy Functional Equation**, whose solutions over $ \mathbb{R} $ are linear functions $ f(x) = kx $, assuming some regularity condition (like continuity, monotonicity, or boundedness). Since the problem does not restrict $ f $, but asks for the **number** of such functions, we must consider all additive functions from $ \mathbb{R} \to \mathbb{R} $.

📌 Under the Axiom of Choice, there are **infinitely many** additive functions (including pathological ones), but if we restrict to **real-valued additive functions** (without further constraints), the number of such functions is uncountably infinite.

📌 However, in the context of a math olympiad, and given the phrasing find the number of functions, we assume the standard assumption of **linear solutions**, i.e., $ f(x) = kx $.

📌 Thus, the number of such functions is equal to the number of real constants $ k $, which is uncountably infinite. But if we are to count **distinct functions**, and assuming only linear solutions are acceptable, the number is infinite.

📌 However, if the problem is interpreted as asking for the number of **continuous** solutions, then the only such functions are linear: $ f(x) = kx $, and there are infinitely many such functions (one for each real $ k $).

📌 Hence, the number of such functions is $ \boxed{\infty} $, assuming no further restrictions.

📌 Question: Let $ x, y, z $ be positive real numbers such that $ x + y + z = 3 $. Find the minimum value of

📌 \frac{x^2}{y + z} + \frac{y^2}{z + x} + \frac{z^2}{x + y}.

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Thus, the square is \(\boxed{50}\).
Factor the expression \( x^4 - 16y^4 \) completely over the real numbe...
Recognize the expression as a difference of squares:
x^4 - 16y^4 = (x^2)^2 - (4y^2)^2 = (x^2 - 4y^2)(x^2 + 4y^2)
Now, note that \(x^2 - 4y^2\) is also a difference of squares:
x^2 - 4y^2 = (x - 2y)(x + 2y)
The term \(x^2 + 4y^2\) cannot be factored over the reals.
Thus, the complete factorization is:
Final answer: \(\boxed{(x - 2y)(x + 2y)(x^2 + 4y^2)}\)
A sequence of five real numbers forms an arithmetic progression, and t...

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