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Can there be two square rational numbers that are equidistant to $1$ i.e. there is a rational number $a \in [0,1]$ such that both $1-a$ and $1+a$ are square rationals?

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Note that two numbers that sum to 2 are equidistant to 1. Now take your favorite Pythagorean triple

$$x^2 + y^2 = z^2$$

and see that

$$\frac{(x - y)^2}{z^2} + \frac{(x + y)^2}{z^2} = 2$$

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Hint : $$1+49 = 2 \times 25$$

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  • $\begingroup$ So $a=1/25$ and you assumed that it has numerator $1$ to check if this would work? Thanks though. I wonder if this is related to a more difficult diophantine equation. $\endgroup$ – quantum Jun 26 at 23:14
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    $\begingroup$ Note: also $ 1+1681=2\times841$ $\endgroup$ – J. W. Tanner Jun 26 at 23:16
  • $\begingroup$ @J.W.Tanner there are many many more including $169+8281 = 529+7921 = 1225+7225 = 2209+6241 = 2 \times 4225$ $\endgroup$ – Henry Jun 26 at 23:21
  • $\begingroup$ @quantum - not quite: but $a=\frac{24}{25}$ works $\endgroup$ – Henry Jun 26 at 23:22

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