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1. Ive been trying to do this problem but have yet to come up with a solution/answer.
Can I have some help on how to solve this problem?

Is there a positive integer k such that k^3+2k^2+2k+1 is a cube number?
2. (Original post by JackSpinner1)
Ive been trying to do this problem but have yet to come up with a solution/answer.
Can I have some help on how to solve this problem?

Is there an interger k such that k^3+2k^2+2k+1 is a cube number?
I assume the question asks positive integers, since otherwise -1 and 0 are obvious solutions
1)consider k^3
2)consider (k+1)^3
3)remember that those two numbers are consecutive cubes
4)look at the expression again
5)win
3. k=-1...

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4. (Original post by apriori1234)
k=-1...

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In the original MAT question it's specified k is positive.
5. (Original post by RichE)
In the original MAT question it's specified k is positive.
Lol I would've expected as much

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6. (Original post by ValerieKR)
I assume the question asks positive integers, since otherwise -1 and 0 are obvious solutions
1)consider k^3
2)consider (k+1)^3
3)remember that those two numbers are consecutive cubes
4)look at the expression again
5)win
Is the answer no because k^3 is the first cube number and (k+1)^3 = k^3+3k^2+3k+1(which is the next cube number) is not equal to the expression stated and so no postive number would satisfy it.
7. (Original post by JackSpinner1)
Is the answer no because k^3 is the first cube number and (k+1)^3 = k^3+3k^2+3k+1 which is not equal to the expression stated and so no postive number would satisfy it.
you tell me
8. You can show that if then unless .
9. (Original post by JackSpinner1)
Is the answer no because k^3 is the first cube number and (k+1)^3 = k^3+3k^2+3k+1(which is the next cube number) is not equal to the expression stated and so no postive number would satisfy it.
Yes, that basically works.

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