since 4x + 1 > 4x and 4x + 1 > x + 1 for x > 0, then by Pythagorean formula
(4x)² + (x + 1)² = (4x + 1)²
16x² + x² + 2x + 1 = 16x² + 8x + 1
x² - 6x = 0
x = 6
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since 4x + 1 > 4x and 4x + 1 > x + 1 for x > 0, then by Pythagorean formula
(4x)² + (x + 1)² = (4x + 1)²
16x² + x² + 2x + 1 = 16x² + 8x + 1
x² - 6x = 0
x = 6