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Show that the equation x^2+y^2=0.999999 has no rational solutions.

  1. Show that if there was a solution, then there must be one of the form x=(a/1000c) and y=(b/1000c) where a,b,c have no divisors greater than 1 in common. Conclude that a solution to the initial equation would also give an integer solution to a^2+b^2=999999^2.
  2. Use congruences to find all possible remainders of a square in division by 7.

3.Find the factorization of 999999 into primes and use it to find a common factor of a,b, and c/

Ben French
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  • the steps here are already fairly detailed, what are you having a problem with exactly? what did you try to solve it? – gt6989b Oct 23 '17 at 15:07
  • I solved part two by evaluating each possible remainder in mod7 then squaring them in mod 7, and was able to factor 9999999 into 3^3 7111337 however im not sure how to solve the first step – Ben French Oct 23 '17 at 15:14

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Here is a general fact, proved here:

Let $r=\frac{p}{q}$, where $p$ and $q$ are integers. Then $r$ can be written as the sum of two rational squares if and only if $pq$ can be written as the sum of the squares of two integers.

Now $ 0.999999 = \dfrac{999999}{1000000} $ and $999999\cdot 1000000 = 2^6×3^3×5^6×7×11×13×37$. Since $7$ appears with odd exponent, $999999\cdot 1000000$ cannot be written as the sum of the squares of two integers.

lhf
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