S \geq 1.

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• S = \frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x},
• given that $ x + y + z = 1 $ and $ x, y, z > 0 $, we apply the **Cauchy-Schwarz Inequality** in the following form:
• Since $ x + y + z = 1 $, we substitute into the inequality:
• To check if this bound is achievable, consider the equality condition in Cauchy-Schwarz: equality occurs when
• Let this common value be $ k $. Then,
• \frac{x^2}{y} = \frac{y^2}{z} \Rightarrow x^2 z = y^3, \quad \frac{y^2}{z} = \frac{z^2}{x} \Rightarrow y^2 x = z^3.

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\( 4^4 = (16)^2 = (-1)^2 = 1 \mod 17 \) → yes\( 5^4 = 625 \). \( 625 \div 17 \): \( 17 \cdot 36 = 612 \), \( 625 - 612 = 13\( 6^4 = (36)^2 \equiv (-5)^2 = 25 \equiv 8\( 7^4 = (49)^2 \equiv (49 - 2\cdot17 = 15)^2 = 225 \equiv 225 - 13\cdot17 = 225 - 221 = 4\( 8^4 = (64)^2 \equiv (64 - 3\cdot17 = 13)^2 = 169 \equiv 169 - 9\cdot17 = 169 - 153 = 16\( 9^4 = (81)^2 \equiv (81 \mod 17 = 13)^2 = 169 \equiv 16\( 10^4 = (100)^2 \equiv (100 - 5\cdot17 = 15)^2 = 225 \equiv 4\( 11^4 = (121)^2 \equiv (121 - 7\cdot17 = 121 - 119 = 2)^2 = 4\( 12^4 = (144)^2 \equiv (144 - 8\cdot17 = 144 - 136 = 8)^2 = 64 \equiv 13\( 13^4 = (169)^2 \equiv (16)^2 = 256 \equiv 256 - 15\cdot17 = 256 - 255 = 1 \) → yes