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LostInParadise's avatar

Care to try this problem with whole numbers?

Asked by LostInParadise (32202points) 3 days ago

Looked at in the right way, this is a fairly easy problem to solve (think in terms of prime numbers).

Given two whole numbers x and y, what are the conditions for xy and x/y both being squares of whole numbers?

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4 Answers

elbanditoroso's avatar

x = 0 y = 0

x = 1 y = 1

LostInParadise's avatar

x/y = 0/0 is undefined
Looking for general case.

ragingloli's avatar

In my mind, it would have to by x=y, so that xy is a square by default, and x/y=1, which is 1²

LostInParadise's avatar

The interesting thing to point out is that xy and x/y are both perfect squares or both not perfect squares, You can see this by doing a prime factorization of x and y. To be perfect squaresm the exponent of each prime factor must be even. The exponent of a prime p in xy is the sum of the prime factors of p for x and y and the exponent of p for x/y is the difference of the exponents. The sum of two numbers is even if and only if the difference is even.

Speaking of perfect squares, next uear, 2025 = 45*2i, is the only year this century that is a perfect square. As the years get higher numbers, such years become less frequent The 36th cintury will be the first not to have any such years.

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