Problem no. 1 from Abhinav

Number Theory Level 3

Does the equation have integer solution?

( a + 1 ) 3 + ( a + 2 ) 3 + … + ( a + 6 ) 3 + ( a + 7 ) 3 = ( b ) 4 + ( b + 1 ) 4 (a+1)^3 + (a+2)^3 + \ldots + (a+6)^3 + (a+7)^3 = (b)^4 + (b+1)^4

No Yes

Abhinav Shripad by Abhinav Shripad , 17, India

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2 solutions

Cantdo Math
May 1, 2020

The LHS must be 0 ( m o d 7 ) 0 (\mod 7) .Since it is the same as 1 3 + . . . 7 3 ( m o d 7 ) 1^3+...7^3 (\mod7) which is 0 if we calculate it.

But it can be easily checked that the RHS can never be 0 m o d 7 0 \mod 7 . Hence,no such ( a , b ) (a,b) exist.

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James Wilson
Dec 16, 2017

The left side is even, and the right side is odd.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Its incorrect, the left side can be odd if a a is odd. The correct solution is taking mod 7 both sides

Shreyansh Mukhopadhyay - 3 years, 3 months ago

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My solution is indeed incorrect. Thank you for pointing that out. Do you intend to add the correct solution or would you like me to correct mine, and give you credit?

James Wilson - 3 years, 3 months ago

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