Don't let you its appearances fool you

n 2 n^{2} + n + 41 is prime for every natural number n.

Is this statement true, or false?

True False

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3 solutions

Santiago Arce
Apr 9, 2018

First we will take a factor of n from the first two terms of the expression. Leaving us with:

n(n+1)+41

We will work with this now to prove several things. The first is that the statement from the problem is false. We can do this through a counter example: substituting n for 41, we have:

41(42)+41 = 41(42+1)

Which is not a prime number, and since 41 is a natural number, then the statement is false . As we can see, this would also be true for n = 40 (Because n + 1 = 41, thus the whole expression is a multiple of 41, and not prime.) With the problem solved, we could also look for the cases in which the expression is not prime. It is easy to see that, since 41 is prime, the only time n(n+1) and 41 are going to have any factors in common is if either n, or n+1 are a multiple of 41. This is why it is a nice problem for beginner mathematicians. Because they won't be able to try and test with different values for n, because the first number that will not make the expression prime is 40.

I believe your second equation should have been 41 ( 42 ) + 41 = 41 ( 42 + 1 ) 41(\color{#D61F06}{42}\color{#333333}{)+41=41(}\color{#D61F06}{42}\color{#333333}{+1)} .

David Stiff - 3 years, 2 months ago

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Yes. Thank you.

Santiago Arce - 3 years, 2 months ago
Edwin Gray
Sep 18, 2018

Let n = 41. Ed Gray

Giorgos K.
Apr 10, 2018

here are the first cases that return false

40 , 41 , 44 , 49 , 56 , 65 , 76 , 81 , 82 , 84 , 87 , 89 , 91 , 96... 40, 41, 44, 49, 56, 65, 76, 81, 82, 84, 87, 89, 91, 96...

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