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I have a sequence of 21 numbers (-10, -9, -8, ..., -1, 0 , 1, 2, ... , 10). I can take any pair $(a, b)$ from the sequence and replace it with pair $((3a - 4b)/5 ), (4a + 3b)/5)$. Is it possible that all numbers in a sequence will be equal after performing some operations?

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    one idea : if operation is convergent and sequence is {a,a,...,a} then (a,a) -> (-a/5, 7a/5) so it implies a=0. – 2017-02-18
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    According to quasi solution since $x^2=\frac{N(N+1)}{3}$ it could be possible for instance for $N=3$ or $N=48$ instead of $N=10$. – 2017-02-18

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Note that after any replacement, all terms of the sequence are still rational numbers.

Suppose in a given step, you replace the pair $\small{\displaystyle{(a, b)}}$ by the pair $\small{\displaystyle{\left(\frac{3a - 4b}{5}, \frac{4a + 3b}{5}\right)}}$.

Observing that

\begin{align*} \left(\frac{3a - 4b}{5}\right)^2 + \left(\frac{4a + 3b}{5}\right)^2 &= \frac{9a^2 - 24ab + 16b^2}{25} + \frac{16a^2 + 24ab + 9b^2}{25}\\[6pt] &=\frac{25a^2+25b^2}{25}\\[6pt] &=a^2 + b^2 \end{align*}

it follows that, after any replacement, the sum of the squares of the terms remains the same.

Suppose that after some number of replacements, the terms are all equal to $x$ say.

Since the sum of the squares must be equal to the original sum of the squares, we get

\begin{align*} &x^2 + x^2 + x^2 + \cdots + x^2 = (-10)^2 + (-9)^2 + (-8)^2 + \cdots + 8^2 + 9^2 + {10}^2\\[6pt] \implies\; &21x^2=2(1^2+2^2+3^2 + \cdots + 10^2)\\[6pt] \implies\; &21x^2=2\left(\frac{(10)(10+1)(2(10)+1)}{6}\right)\\[6pt] \implies\; &21x^2=(10)(11)(7)\\[6pt] \implies\; &x^2 = {\small{\frac{110}{3}}}\\[6pt] \end{align*}

contradiction, since $\large{\frac{110}{3}}$ is not the square of a rational number.

It follows that it's not possible, after some number of replacements, to get all equal terms.

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    should divide by 6 so $3x^2=110$ but conclusion is the same. – 2017-02-18
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    @zwim: Thanks. Now fixed. – 2017-02-18