BMO1_2019_2

BMO1_2019_1

I claim that all odd integer values work for a_1.
To prove this, let’s first factorize the given equation.
Looking at the 4's we immediately think of squaring a 2.

This method proves to work, as you can indeed factor the given recursive formula as:
(2a_{n+1}-a_n)^2=1

Taking the square root we get:
2a_{n+1}-a_n = \pm 1

Isolating a_{n+1} we can derive two possible values for a_n.

a_{n+1}=\frac{a_n \pm 1}{2}.
From here we can see that for a_{n+1} to be an integer, a_n has to be odd, in order to be able to divide by 2.

Now we prove that all odd values work for a_1 and that we can continue to build an integer sequence.

Note that the two values, \frac{a_n-1}{2} and \frac{a_n+1}{2} are 1 apart from each other.
This means that they are consecutive integers, so one will be odd while the other, even.
Eureka!
To continue our sequence we can keep on choosing the odd value (out of these two)!
Thus our claim holds and we are done.

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