The sum is simultaneously odd and even

Number Theory Level 2

1 + 3 + 5 + 7 + ⋯ + ( 2 m + 1 ) = 2 + 4 + 6 + 8 + ⋯ + ( 2 n ) 1 + 3 + 5 + 7 + \cdots + (2m + 1) = 2 + 4 + 6 + 8 + \cdots + (2n) Is it possible to have an equality where the left-hand side is the sum of all odd positive integers up to a certain point, as shown, and similarly the right-hand side is the sum of even positive integers up to a certain point?

No Yes

Pi Han Goh by Pi Han Goh , 30, Malaysia

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3 solutions

Otto Bretscher
Dec 10, 2018

The left-hand side is a perfect square, ( m + 1 ) 2 (m+1)^2 , while the right-hand side, n ( n + 1 ) n(n+1) isn't since n 2 < n ( n + 1 ) < ( n + 1 ) 2 n^2<n(n+1)<(n+1)^2 .

Clever problem!

5 Helpful 1 Interesting 2 Brilliant 0 Confused
Ritabrata Roy
Jan 18, 2019

LHS=(2m+1)^2=odd number 

 RHS=n(n+1)=even number

So, there is no solution.

0 Helpful 0 Interesting 1 Brilliant 0 Confused

Thats what I did

Sachin Mital - 2 years, 1 month ago
Sachin Mital
Apr 26, 2019

Consider 2+4+6, this is equal to: 1+3+5+3(1). This can be explained in a more general way: for n terms in the RHS, it is equal to n terms in the LHS plus n. Since the n+1th term will be 2n+1, there can never be a case where LHS=RHS.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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