- Thread starter PapaKurke
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There is no reason for you to think of yourself as dumb!

As Jomo points out \(\displaystyle n=1\) does work. So you should have tried that.

But I cannot see this as an eighth grade question. look at this.

None of the next 49 odd powers work. If I were you, I would go to the authority of eight grader and demand to know why this question was given. I would happily give my assessment of this question.

Problems in number theory are seldom easy, and I have no reason to think that it is useful to pose such problems to 8th graders. For example, your result that any such number is divisible by 3 is very nice. By the binomial theorem,

\(\displaystyle 2^n + 3^n + 4^n = (3 - 1)^n + 3^n + (3 + 1)^n =\)

is certainly divisible by 3 because

\(\displaystyle (-\ 1)^{(2k-1)} * \dbinom{2k-1}{0} * 3^0 * 1^{(2k-1)} + \dbinom{2k-1}{0} * 3^0 * 1^{(2k-1)} = 0.\)

Well done. Unfortunately, few 8th graders know the binomial theorem so I doubt that was a possible approach that they were intended to explore.

One possibility is that n = 1 is the only answer. Shades of Fermat's Last Theorem seem to lurk. My suggestion is to use a computer to check whether any n works in the range of numbers reachable by your computer. If one does, think about mathematical induction.

EDIT: Check to see whether a bound was placed on n in the original problem. That makes the problem rather foolish, but**THEN** it certainly is within the capabilities of someone in eighth grade.

\(\displaystyle 2^n + 3^n + 4^n = (3 - 1)^n + 3^n + (3 + 1)^n =\)

is certainly divisible by 3 because

\(\displaystyle (-\ 1)^{(2k-1)} * \dbinom{2k-1}{0} * 3^0 * 1^{(2k-1)} + \dbinom{2k-1}{0} * 3^0 * 1^{(2k-1)} = 0.\)

Well done. Unfortunately, few 8th graders know the binomial theorem so I doubt that was a possible approach that they were intended to explore.

One possibility is that n = 1 is the only answer. Shades of Fermat's Last Theorem seem to lurk. My suggestion is to use a computer to check whether any n works in the range of numbers reachable by your computer. If one does, think about mathematical induction.

EDIT: Check to see whether a bound was placed on n in the original problem. That makes the problem rather foolish, but

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Thank you all for your replies. No, there is no bound on n, and no, this is definitely not a simple problem. I have managed, however, to solve it. The answer came to me, surprisingly enough, in my sleep.

Suppose n>1 , then 4|(2^n) and 4|(4^n) . Since n is odd and 3 = (-1) mod 4 then it holds 3^n = (-1) mod 4. We have therefore proven that for all n>1 the expression gives a remainder of 3 when diving by 4.

Looking at any perfect square at all we see that they all give a remainder of either 0 or 1 with 4:

(0)^2 = 0 mod 4

(1)^2 = 1 mod 4

(2)^2 = 0 mod 4

(3)^2 = 1 mod 4.

Which means that for n>1 the given expression is not a perfect square.

We therefore only need to check n=1 which, conveniently, is a solution, so the only solution is n=1.

I once again thank you for all your replies and exclaim once more how flawed our educational system is giving this to 8th graders. I am a 10th grader and could barely come up with this. Perhaps there is a more elegant solution I am simply failing to see?

Edit #1: Commenting further on the problem. For even n there are, I would at least believe, more solutions as the given expression gives a remainder of 1 with 4 bringing the possibility of the expression being a perfect square back.

Edit #2: Giving the problem for even n another good look we see that the expression is (-1)^2n + (0)^2n + (1)^2n = 2 mod 3 , and since no squares give a remainder of 2 when divided by 3 (easily proven similarly to how no squares give a remainder of 3 with 4) there are no such even numbers n.

Suppose n>1 , then 4|(2^n) and 4|(4^n) . Since n is odd and 3 = (-1) mod 4 then it holds 3^n = (-1) mod 4. We have therefore proven that for all n>1 the expression gives a remainder of 3 when diving by 4.

Looking at any perfect square at all we see that they all give a remainder of either 0 or 1 with 4:

(0)^2 = 0 mod 4

(1)^2 = 1 mod 4

(2)^2 = 0 mod 4

(3)^2 = 1 mod 4.

Which means that for n>1 the given expression is not a perfect square.

We therefore only need to check n=1 which, conveniently, is a solution, so the only solution is n=1.

I once again thank you for all your replies and exclaim once more how flawed our educational system is giving this to 8th graders. I am a 10th grader and could barely come up with this. Perhaps there is a more elegant solution I am simply failing to see?

Edit #1: Commenting further on the problem. For even n there are, I would at least believe, more solutions as the given expression gives a remainder of 1 with 4 bringing the possibility of the expression being a perfect square back.

Edit #2: Giving the problem for even n another good look we see that the expression is (-1)^2n + (0)^2n + (1)^2n = 2 mod 3 , and since no squares give a remainder of 2 when divided by 3 (easily proven similarly to how no squares give a remainder of 3 with 4) there are no such even numbers n.

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Yes, but the original problem restricted n to odd integers.It is only divisible by 3 for odd values of n

Note that 3 does not divide (2^2 + 3^2 + 4^2) =29

Can n be even and we get the desired results?

The problem was given to 8th graders as a preparation for the local city competition along with other banal problems such as solving a system of linear equations and solving the absolute simplest of geometry problems (such as proving a quadrilateral is a parallelogram if its opposing angles are equal and a single pair of its sides are parallel). This one definitely stuck out, as there is no business for congruencies to be something 8th graders should worry about at this level. Formally, they should appear only at national junior olympiad level.