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Posted on 2009-04-23

Find an integer n so that n, n + 1, n + 2, n + 3 are all divisible by perfect

squares.

squares.

17 Comments

The smallest is n = 157² = 24649

Others are 193², 239², 257², 293², 437², 607², 643², 707², 743²

57121 = 239² 57122 = 338 x 13² 57123 = 6347 x 3² 57124 = 14281 x 2²

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.

http://en.wikipedia.org/wi

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.

Another approach is described here: http://www.marmet.org/loui

242 = 2*11^2

243 = 27*3^2

244 = 61*2^2

245 = 5*7^2

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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