# Proving theorems and special cases (Part 1): Is n^2-n+41 always prime?

In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:

Is it ever possible to prove a statement or theorem by proving a special case of the statement or theorem?

Usually, the answer is no… even checking many special cases of a conjecture does not mean that the conjecture is correct.

The following example probably appears in every textbook that I’ve seen that handles mathematical induction to convince students that checking even many special cases of a conjecture is not sufficient for proving the conjecture.

Conjecture: If $n \ge 1$ is a positive integer, then $n^2 - n + 41$ is a prime number.

Is this true? Well, let’s start checking:

If $n = 1$, then $n^2 - n + 41 = 1^2 - 1 + 41 = 41$, which is a prime number.

If $n = 2$, then $n^2 - n + 41 = 2^2 - 2 + 41 = 43$, which is a prime number.

If $n = 3$, then $n^2 - n + 41 = 3^2 - 3 + 41 = 47$, which is a prime number.

If $n = 4$, then $n^2 - n + 41 = 4^2 - 4 + 41 = 53$, which is a prime number.

If $n = 5$, then $n^2 - n + 41 = 5^2 - 5 + 41 = 61$, which is a prime number.

If $n = 6$, then $n^2 - n + 41 = 6^2 - 6 + 41 = 71$, which is a prime number.

If $n = 7$, then $n^2 - n + 41 = 7^2 - 7 + 41 = 83$, which is a prime number.

If $n = 8$, then $n^2 - n + 41 = 8^2 - 8 + 41 = 97$, which is a prime number.

If $n = 9$, then $n^2 - n + 41 = 9^2 - 9 + 41 = 113$, which is a prime number.

If $n = 10$, then $n^2 - n + 41 = 10^2 - 10 + 41 = 131$, which is a prime number.

If $n = 11$, then $n^2 - n + 41 = 11^2 - 11 + 41 = 151$, which is a prime number.

If $n = 12$, then $n^2 - n + 41 = 12^2 - 12 + 41 = 173$, which is a prime number.

If $n = 13$, then $n^2 - n + 41 = 13^2 - 13 + 41 = 197$, which is a prime number.

If $n = 14$, then $n^2 - n + 41 = 14^2 - 14 + 41 = 223$, which is a prime number.

If $n = 15$, then $n^2 - n + 41 = 15^2 - 15 + 41 = 251$, which is a prime number.

If $n = 16$, then $n^2 - n + 41 = 16^2 - 16 + 41 = 281$, which is a prime number.

If $n = 17$, then $n^2 - n + 41 = 17^2 - 17 + 41 = 313$, which is a prime number.

If $n = 18$, then $n^2 - n + 41 = 18^2 - 18 + 41 = 347$, which is a prime number.

If $n = 19$, then $n^2 - n + 41 = 19^2 - 19 + 41 = 383$, which is a prime number.

If $n = 20$, then $n^2 - n + 41 = 20^2 - 20 + 41 = 421$, which is a prime number.

If $n = 21$, then $n^2 - n + 41 = 21^2 - 21 + 41 = 461$, which is a prime number.

If $n = 22$, then $n^2 - n + 41 = 22^2 - 22 + 41 = 503$, which is a prime number.

If $n = 23$, then $n^2 - n + 41 = 23^2 - 23 + 41 = 547$, which is a prime number.

If $n = 24$, then $n^2 - n + 41 = 24^2 - 24 + 41 = 593$, which is a prime number.

If $n = 25$, then $n^2 - n + 41 = 25^2 - 25 + 41 = 641$, which is a prime number.

If $n = 26$, then $n^2 - n + 41 = 26^2 - 26 + 41 = 691$, which is a prime number.

If $n = 27$, then $n^2 - n + 41 = 27^2 - 27 + 41 = 743$, which is a prime number.

If $n = 28$, then $n^2 - n + 41 = 28^2 - 28 + 41 = 797$, which is a prime number.

If $n = 29$, then $n^2 - n + 41 = 29^2 - 29 + 41 = 853$, which is a prime number.

If $n = 30$, then $n^2 - n + 41 = 30^2 - 30 + 41 = 911$, which is a prime number.

If $n = 31$, then $n^2 - n + 41 = 31^2 - 31 + 41 = 971$, which is a prime number.

If $n = 32$, then $n^2 - n + 41 = 32^2 - 32 + 41 = 1033$, which is a prime number.

If $n = 33$, then $n^2 - n + 41 = 33^2 - 33 + 41 = 1097$, which is a prime number.

If $n = 34$, then $n^2 - n + 41 = 34^2 - 34 + 41 = 1163$, which is a prime number.

If $n = 35$, then $n^2 - n + 41 = 35^2 - 35 + 41 = 1231$, which is a prime number.

If $n = 36$, then $n^2 - n + 41 = 36^2 - 36 + 41 = 1301$, which is a prime number.

If $n = 37$, then $n^2 - n + 41 = 37^2 - 37 + 41 = 1373$, which is a prime number.

If $n = 38$, then $n^2 - n + 41 = 38^2 - 38 + 41 = 1447$, which is a prime number.

If $n = 39$, then $n^2 - n + 41 = 39^2 - 39 + 41 = 1523$, which is a prime number.

If $n = 40$, then $n^2 - n + 41 = 40^2 - 40 + 41 = 1601$, which is a prime number.

Okay, a show of hands… did anyone actually carefully read and check the above 40 lines? I didn’t think so. The point is that the proposition works for $n = 1, 2, 3, \dots, 40$. By about $n = 4$, or so, a student (who didn’t already know the answer) actually did the above work would begin thinking, “Wow, this probably is correct for any value of $n$.”

Of course, the catch happens at $n = 41$:

If latex n = 41, then $n^2 - n + 41 = 41^2 - 41 + 41 = 41^2 = 41 \times 41$,

which is not a prime number.

All this to say, seeing a trend for the first few special cases… or the first few dozen special cases… does not necessarily mean that the trend will continue.

## One thought on “Proving theorems and special cases (Part 1): Is n^2-n+41 always prime?”

This site uses Akismet to reduce spam. Learn how your comment data is processed.