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Last edited by tony123 (2007-09-08 04:29:43)
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Just to point out, that recurrence relation should really define that k_2 = 30 as well.
It's fairly easy to prove that all terms in this sequence are multiples of 10.
That means that multiplying two consecutive terms in the sequence together and then by 5 will always get you a multiple of 500.
Therefore, 1 + 5[K_nK_(n+1)] will always have its last 3 digits as 001 or 501.
If you can prove that no squares end in those digits then you'll have proved that no n will yield a perfect square.
Of course, it's possible that there are squares with that ending and that I'm on completely the wrong track.
Edit: And there are. Ah well. Hopefully I've given someone else inspiration.
Why did the vector cross the road?
It wanted to be normal.
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A perfect square ending in 1 must be the square of a number ending in 1 or 9, i.e. it must be of the form
If its to end in 001 or 501, N must be a multiple of 25. Hence, the problem comes down to finding n such that K[sub]n[/sub] is a multiple of 50, or otherwise proving that no such n exists.
NB:
If N is odd, both 62500N[sup]2[/sup] and 500N end in 500, so their sum/difference must end in 000. And if N is even, both 62500N[sup]2[/sup] and 500N end in 000. Hence the square of (250N±1) always ends in 001.
Last edited by JaneFairfax (2007-09-08 10:41:54)
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