x^2 + x + 41?

Question

Euler's formula produces primes for many values of X, but it does not work for all of them. Find the first value of X for which the formula fails. (hint: try multiples of ten) please help me i went to 1000 and still couldn't find anything.

Answer 1

40

Answer 2

It fails when n=40

40^2+40+41 = 1681 = 41^2

Answer 3

Sorry, it does fail at 40.

40² = 1600

40 + 41 = 81

Total: 1681 = 41².

Note: It also fails at 41: 41² + 41 + 41 is divisible by 41.

As to why this gives primes for x =1, ... , 39,

there is a fascinating theory based on the fact

that the quadratic field Q(√-163 ) has unique

factorisation. You might want to consult Wikipedia

for more information.

Answer 4

It fails at x=40 and x=41.



Of course if fails at x=40:



x^2 + x + 41 = 40^2 + 40 + 41 = 1600 + 40 + 41 = 1681



And if you check with a calculator, 1681 = 41^2



Therefore, it is a composite number (as it has factors 1, 41 and itself). Hence, Euler's formula fails at x=40 itself.

Answer 5

It fails at 41. The polynomial = 1681 which is the square of 41.

Answer 6

40 cause

40*40=1600

1600+40=1640

1640+41=1681

1681/41=41

Answer 7

Damn it woman, do your homework!