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📰 \(\sqrt{289} = 17\), and \(17 \times 17 = 289\).
📰 Thus, \(2023 = 7 \times 17^2\).
📰 The divisors of 2023 are \(1, 7, 17, 49, 119, 289, 2023\). The largest divisor is 2023 itself. If \(d = 2023\), then \(x + y = 1\), which implies one of \(x\) or \(y\) is 0, contradicting \(x, y\) being positive integers. So, the largest feasible \(d\) is 289.
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