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# Arithmetic series watch

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1. Prove that if one arithmetic series has 1 square and one cube, it will has one term is the power of six of an integer.
2. (Original post by BCHL85)
Prove that if one arithmetic series has 1 square and one cube, it will has one term is the power of six of an integer.
Without any loss of generality one can start the arithmetic progression from the square. Call this first term a^2 and call the common difference d.

Then for some k and some b

a+kd=b^3

So, squaring

a^2 + (2ka+dk^2)d =b^6

which is the (2ka+dk^2-1)th term in the AP.
3. But it isnt the same AP as the one with the cube term. They have different first terms.
4. (Original post by JamesF)
But it isnt the same AP as the one with the cube term. They have different first terms.
You're right - wasn't thinking!
5. (Original post by JamesF)
But it isnt the same AP as the one with the cube term. They have different first terms.
What do you mean? The question says one AP.
6. (Original post by J.F.N)
What do you mean? The question says one AP.
I started one of my APs with term a, one with a^2 - that was my error.
7. There are some questions relate to that
- If a arithmetic series has one square. Prove that there's another one in this series.
- 2nd is like I posted.
- If a arithmetic series has 3 terms are the powers of 2, 3, 5. Prove that there's a term is power of 2.3.5 = 30.
- If it has n terms are the powers of n numbers, every couples are prime number with each other. Prove that there's a term is the power of multiplication of these numbers.
Woa ... hard to express the general question.

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