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# divisible by perfect squares

Posted on 2009-04-23
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Find an integer n so that n, n + 1, n + 2, n + 3 are all divisible by perfect
squares.

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Question by:AlephNought

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Any number is divisible by 1 (which is a square number), so any value for n should work, no ?
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Is this homework or a puzzle?
What is the title of the chapter that this problem comes from?
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2^2 | 350
3^2 | 351
2^2 | 352

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I have found several:

The smallest is   n =  157² = 24649

Others are     193²,  239²,  257²,  293²,  437²,  607²,   643²,   707²,  743²
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Mmm, we seem to be using different definitions of "perfect square". AlephNought, what's your definition of a "perfect square" ?
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Assisted Solution

24649 = 157²      24650 = 986 x 5²      24651 = 2739 x 3²      24652 = 6163 x 2²

57121 = 239²      57122 = 338 x 13²    57123 = 6347 x 3²      57124 = 14281 x 2²
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d-glitch, I take it your definition of "perfect square" is the square of a prime number ?

Another definition of "perfect square" is that it's the sum of two other squares (a la 5² = 4² + 3²).

Yet another definition of "perfect square" is simply a square number (which is the definition I used).

There might even be more ... but those seem to be the most common.
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If the sqrt(N) is an integer (not necessarily a prime), then N is a perfect square.

http://en.wikipedia.org/wiki/Square_number

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Assisted Solution

Yep, that's what I used too. So, N=1 applies :

1 = 1²
2 = 1² * 2
3 = 1² * 3
4 = 2²
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Expert Comment

Maybe, but that seems like a trivial case.
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;) Sure. But it's a perfectly valid answer, assuming that's the definition of "perfect square" that AlephNought had in mind.
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If this is a trick question like  "Who is buried in Grant's Tomb?"  Then your first post is correct.

Any and all sequences of four integers are divisible by 1².

But I expect there may have been some sort of error in the transcription of the problem.
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Accepted Solution

Interesting problem. I thought you could use the Mobius function and look for 4 consecutive numbers where mu(n_i) = 0 but I'm not sure that path leads anywhere.

Another approach is described here: http://www.marmet.org/louis/sqfgap/ and you'd be looking for numbers with L>=4. According to that site, the smallest such number is 242

242 =  2*11^2
243 = 27*3^2
244 = 61*2^2
245 = 5*7^2

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Sounds plausible :)
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>> Sounds plausible :)

That was in response to d-glitch's comment btw
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I agree: gatorvip does seem to have a more elegant solution.
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Unfortunately, I wouldn't call it a solution, but more of a method to find numbers like that. When L>=4 then you have L-3 consecutive possibilities.

If L=4, the smallest such number is 242. Then the rest of the numbers in the series are 3174, 3750, ...
If L=5, the smallest such number is 844, so both 844 and 845 work. The next numbers are 1680 and 1681, which are both solutions, etc.
If L=6, the smallest such number is  22020, so you get valid answers in 22020, 22021, 22022.

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