# Thread: For Theatregirl -- math problem

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## For Theatregirl -- math problem

Is the expression x&#178; + x + 41 prime for all non-negative integers x?

0&#178; + 0 + 41 = 41 (prime)
1&#178; + 1 + 41 = 43 (prime)
2&#178; + 2 + 41 = 47 (prime)
3&#178; + 3 + 41 = 53 (prime)
4&#178; + 4 + 41 = 61 (prime)
5&#178; + 5 + 41 = 71 (prime)

etc. (?)

Note: this sequence can also be obtained by starting with 41, then add on 2, then add on 4, then 6, then 8,...

Extra credit (+3 GOE): Can you settle this question without a calculator?

MM

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## math

seems to be all prime. if you look at the difference between the results in the numbers, it is all even intergers of 2, 4, 6, 8 etc

out of curosity, where do you get stuff like this? are you a math teacher?

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Yes, I am a mathematics professor.

This problem was invented by the prolific Swiss mathematician Leonard Euer in the eighteenth century. There was quite a bit of interest in those days in studying the distribution of primes among the integers. People tried to invent polynomials that generated all primes, and Euler's try, with the polynomial x² + x + 41, was one of the simplest.

What about 25² + 25 + 41? Still prime? How about 38² + 38 + 41?

In 1752 Goldbach proved that no polynomial generates all primes (in particular, this one doesn't). What number should you try to get an answer that is not prime?

BTW, want to make a million dollars? That's the prize offered for anyone who can prove the Goldbach Conjecture: Every even number greater that 4 can be written as the sum of two primes.

MM

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Originally Posted by Mathman

BTW, want to make a million dollars? That's the prize offered for anyone who can prove the Goldbach Conjecture: Every even number greater that 4 can be written as the sum of two primes.
How high do you have to go? Does infinity have to be in the equation? Who's offering the prize? It could pay for my dissertation research!

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Originally Posted by Mathman
Yes, I am a mathematics professor.
...........and all time sweetheart too. (Dee running away from Mrs. Math)

Dee

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Thank you! Using the completely non scientific method of guess and check (and I list of the 10,000 smallest primes, but no calculator) I have determined that the answer is 40. 40x40=1600, 1600+40=1640, 1640+41=1681=41^2. Now, I'm sure that there is a more mathmatically sound way to figure this out, but I am proud of myself for getting an answer.

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You might have noticed first that 41 is an obvious counterexample, because 41² + 41 + 41 = 41*(41+1+1).

Then you might have guessed at 40 by thinking of the function as being the same as x*(x+1) + 41, so when x = 40 we have 40*41+41, again with a common factor of 41.

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Originally Posted by Alsace
How high do you have to go? Does infinity have to be in the equation? Who's offering the prize? It could pay for my dissertation research!
The prize was offered by the British publishing house Faber to publicize a novel that they had coming out about a crazy guy who devoted his life to working on this problem (Uncle Petros and the Goldberg Conjecture.)

Yes, you have to "go on to infinity" in the sense that you have to give a general proof that will work for all numbers. After all, if you just checked it for the first million even numbers, there is always the possibility that it would fail for the million and first.

People have tried to investigate this by computer, and have checked all numbers up to 300,000,000,000,000,000 (why? ask Uncle Petros! ), without finding a counterexample.

MM

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OK, TG, are you ready for the next one?

Everybody knows that 3&#178; + 4&#178; = 5&#178;, right?

It is not quite so well known that 3&#179; + 4&#179; +5&#179; = 6&#179;.

But suppose we start with 1. Is there any solution in positive integers k and n to the Diophantine equation

1^k + 2^k + 3^k + ... + n^k = (n+1)^k ?

(This is called the Erdos-Moser equation.)

MM

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MM, we love it when you talk numbers. If you had rattled this stuff off to Michelle she would be in Detroit now.

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Originally Posted by Mathman
But suppose we start with 1. Is there any solution in positive integers k and n to the Diophantine equation

1^k + 2^k + 3^k + ... + n^k = (n+1)^k ?
How about k = 1, n = 2
1^1 + 2^1 = 3^1 (1+2=3)

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Originally Posted by Ptichka
How about k = 1, n = 2
1^1 + 2^1 = 3^1 (1+2=3)
That is the only known solution.

As for the possibility of solutions for n>1, the best result to date is that there is no solution with fewer than 10^(9000000) terms.

http://www.ams.org/mcom/2000-69-229/...99-01088-1.pdf

(Theorem 2, page 410. The authors of this paper are three students of mine, and they obtained this result while still undergraduates.)

MM

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