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No. If one of them were a square number, then its square root must end in 1 or 9 (as this is the only way to make the final digit a one). So the square root is of the form \(10n\pm1\).

If \(10(10n^2\pm2n)+1\) is of the form 111...1, then \(10n^2\pm2n\) is also of the form 111...1 (as it has just had the final 1 taken off). But \(10n^2\pm2n\) is even and 111...1 is odd, so this is not possible.

$$111...1=(10n\pm1)^2$$$$=100n^2\pm20n+1$$ $$=10(10n^2\pm2n)+1$$