ATQ, Let the two integers be ‘x’ and ‘y’ where x > y. Then, according to the question, x × y = x × (1/y) × 121 Or, xy = (121x/y) Or, 121x = xy2 So, y = √121 = 11 {since, ‘x’ and ‘y’ are positive integers} So, x = 28 – 11 = 17 Therefore, x2 = 172 = 289
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