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 Integer Divisibility Dilemma II (Posted on 2014-05-22)
Given that n is an integer, what are the values of n for which n2 - 17 is divisible by 5n+33?

Prove that there are no others.

 No Solution Yet Submitted by K Sengupta Rating: 3.0000 (1 votes)

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 All the math | Comment 5 of 6 |
Here's a worked out solution, by completing the square.

If 5n + 33 divides n^2 - 17, then it also divides

25*(n^2 - 17)
= 25n^2 - 425
= (25n^2 - 1089) + 664
= (5n + 33)*(5n - 33) + 664

Note that 25 is relatively prime to 5n + 33, so this is the case if and only if 5n + 33 divides 664.

664 = 2*2*2*83,

so 5n + 33 must equal 1 or 2 or 4 or 8 or 83 or 166 or 332 or 664,
or -1 or -2 or -4 or -8 or -83 or -166 or -332 or -664

Clearly, if 5n + 33 is positive in ends in 3 or 8.
if 5n + 33 is negative it ends in 2 or 7

so, the only solutions are 5n + 33 = 8 or 83 or -2 or -332
In other words, n = -5 or 10 or -7 or -73

Edited on May 23, 2014, 11:03 pm
 Posted by Steve Herman on 2014-05-23 22:46:01

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