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Equation with factorial (Posted on 2018-10-20) Difficulty: 2 of 5
Solve in integers for x and y: 6(x! + 3) = y^2 + 5

No Solution Yet Submitted by Ady TZIDON    
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soln Comment 1 of 1
x must be less than 5 (it is from choices: 0,1,2,3,4). Why? 
x>=5 --> x! is a multiple of 10, and so is 6(x!). The equation:
6(x!)=y^2 -13 means that for x>=5, y^2 must end in 3, which is impossible for integers.
What works is (2,5) and (3,7) and also (2,-5) and (3,-7) 

Edited on October 21, 2018, 3:32 am
  Posted by Steven Lord on 2018-10-20 11:09:48

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