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How many distinct pairs of perfect squares differ by 35? (the pair $a, b$ is the same as the pair $b, a$.)?

User Zorg
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2 Answers

3 votes

Answer:

2

Explanation:

User Fran Marzoa
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35 has 4 divisors, hence two factor pairs: 1*35 and 5*7. Each corresponds to a set of perfect squares that differ by 35

One pair is ((35±1)/2)^2 = {17^2, 18^2} = {289, 324}
The other is ((7±5)/2)^2 = {1^2, 6^2} = {1, 36}
User Cih
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