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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 $x=(\dfrac{a}{1000c})$ and $y=(\dfrac{b}{1000c})$ 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 $c$.

Thirdy Yabata
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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

1 Answers1

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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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