Can the 2008
th repunit be a perfect square in any integer base x ≥ 2?
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Submitted by K Sengupta
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Rating: 4.0000 (2 votes)
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Solution:
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Let us assume that there exists an integer base x>=2, for which the 2008th repunit is a perfect square.
Then. (x^2008 - 1)/(x-1) is a perfect square.
We substitute (p, q, r) = {x^1004 + 1, x^502 + 1, (x^502 - 1)/(x-1)}
Then, we must have:
(x^2008 - 1)/(x-1) = pqr
Now q and r divide p-2, while r divides q-2.
Accordingly, it follows that the gcd of any two of p, q and r is at most 2.
So, the product pqr is a perfect square if p, r and r are either squares or doubles of squares.
It is clearly observed that neither p nor q can be squares of positive integers.
Accordingly, each of p and q must correspond to twice the square of positive integers.
Therefore, we must have:
4pq = 4*x^1506 + 4*x^1004 + 4*x^502 + 4 must be a perfect square.
However, in that situation:
(2*x^753 + x^251)^2
< 4*x^1506 + 4*x^1004 + 4*x^502 + 4
< (2*x^753 + x^251 + 1)^2
This implies that 4pq is not a perfect square, which leads to a contradiction.
Consequently, the 2008th repunit never be a perfect square in any integer base x ≥ 2.
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