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#1 2007-09-08 04:28:40

tony123
Member
Registered: 2007-08-03
Posts: 228

sequence


sequence 20,30,70,180,470,1230,...
find all natural numbers
such that
is  a perfect squer

Last edited by tony123 (2007-09-08 04:29:43)

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#2 2007-09-08 09:09:44

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: sequence

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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#3 2007-09-08 10:31:56

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: sequence

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 it’s 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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