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1 solution
HINT: Prove 𝑥 cannot be more than 3. SOLUTION:For 𝑥=𝑦, we get (𝑥,𝑦)=(3,3). For 𝑥<𝑦, suppose 3≤𝑥⇒3𝑦≤𝑥𝑦=𝑥+𝑦+(𝑥,𝑦)<𝑦+𝑦+𝑦=3𝑦, which is contradiction. Hence either 𝑥=2 or 1. 𝑥=1 does not give a solution. 𝑥=2⇒𝑦=2+(2,𝑦). Since (2,𝑦) is either 1 or 2, we get the solutions as (𝑥,𝑦)=(2,3) or (2,4). Hence there are total three solutions.