**Hello Codeforces,**↵
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This is mathematical proof that the GCD between the sum of even indices and the sum of odd indices of a consecutive set of integers is always greater than one.↵
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**Note:**↵
A special case for this problem is when the size of the set is <= 2 then the GCD will be equal to 1, So I assume that the given set size is greater than 2.↵
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And also We assume X >= 1.↵
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**First let's define our problem:**↵
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Given a consecutive set of integers: X, X + 1, X + 2, X + 3,.....↵
~~~~~↵
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We want to prove that if we took the sum of the even indices (assuming zero-based indexing):↵
`sumOfEven = X + X + 2 +....`↵
And took the sum of the odd indices:↵
`sum of odd = X + 1 + X + 3 +...`↵
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`Then GCD(sum of even, sum of odd) > 1 always.`↵
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So let's assume C1 will be the number of even indices in our set and C2 will be the number of odd indices in our set.↵
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`sum of even = C1*X + C1*(C1 - 1) = C1*X + C1^2 - C1`↵
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`sum of odd = C2*X + C2^2`↵
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**Now we have two cases:**↵
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**First if C1 = C2:**↵
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`sum of even = C1*X + C1^2 - C1 = C1(X + C1 - 1)`↵
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`sum of odd = C1*X + C1^2 = C1(X + C1)`↵
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We can notice that the GCD will be C1 and since the size of the set size is > 2 then C1 >= 2, the GCD will always be greater than one.↵
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**Second if C1 != C2:**↵
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Then we can notice that C1 = C2 + 1 because if the size is even then C1 must be equal to C2 otherwise C1 will be odd and will be equal to C2 + 1.↵
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`sum of even = C1(X + C1 - 1)`↵
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`sum of odd = (C1 - 1)(X + C1 - 1)`↵
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We can notice that both of the sums are divisible by `X + C1 - 1` then we know that the GCD will be at least `X + C1 - 1` and because if `C1 != C2` and `the size of the set is > 2` then `C1 >= 2` and because `X >= 1` then `X + C1 - 1 >= 2` which implies that GCD between the two sums will be always > 1.↵
↵
↵
↵
↵
↵
This is mathematical proof that the GCD between the sum of even indices and the sum of odd indices of a consecutive set of integers is always greater than one.↵
↵
**Note:**↵
A special case for this problem is when the size of the set is <= 2 then the GCD will be equal to 1, So I assume that the given set size is greater than 2.↵
↵
And also We assume X >= 1.↵
↵
**First let's define our problem:**↵
↵
↵
~~~~~↵
Given a consecutive set of integers: X, X + 1, X + 2, X + 3,.....↵
~~~~~↵
↵
↵
We want to prove that if we took the sum of the even indices (assuming zero-based indexing):↵
`sumOfEven = X + X + 2 +....`↵
And took the sum of the odd indices:↵
`sum of odd = X + 1 + X + 3 +...`↵
↵
`Then GCD(sum of even, sum of odd) > 1 always.`↵
↵
So let's assume C1 will be the number of even indices in our set and C2 will be the number of odd indices in our set.↵
↵
`sum of even = C1*X + C1*(C1 - 1) = C1*X + C1^2 - C1`↵
↵
`sum of odd = C2*X + C2^2`↵
↵
**Now we have two cases:**↵
↵
**First if C1 = C2:**↵
↵
`sum of even = C1*X + C1^2 - C1 = C1(X + C1 - 1)`↵
↵
`sum of odd = C1*X + C1^2 = C1(X + C1)`↵
↵
We can notice that the GCD will be C1 and since the size of the set size is > 2 then C1 >= 2, the GCD will always be greater than one.↵
↵
**Second if C1 != C2:**↵
↵
Then we can notice that C1 = C2 + 1 because if the size is even then C1 must be equal to C2 otherwise C1 will be odd and will be equal to C2 + 1.↵
↵
↵
`sum of even = C1(X + C1 - 1)`↵
↵
`sum of odd = (C1 - 1)(X + C1 - 1)`↵
↵
We can notice that both of the sums are divisible by `X + C1 - 1` then we know that the GCD will be at least `X + C1 - 1` and because if `C1 != C2` and `the size of the set is > 2` then `C1 >= 2` and because `X >= 1` then `X + C1 - 1 >= 2` which implies that GCD between the two sums will be always > 1.
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