Square, square
Randomly choose a positive integer less than and call the number
Then what is the probability that we can choose a positive integer such that where only one of and counts for each
Give your answer in percentage to the nearest integer.
The answer is 100.
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1 solution
We claim that every natural number n can be written in this form. The key idea is to notice that for any natural number m we have
m 2 − ( m + 1 ) 2 − ( m + 2 ) 2 + ( m + 3 ) 2 = m 2 − ( m 2 + 2 m + 1 ) − ( m 2 + 4 m + 4 ) + m 2 + 6 m + 9 = 4
so by
0 = − 1 2 − 2 2 + 3 2 − 4 2 + 5 2 + 6 2 − 7 2
1 = 1 2
2 = − 1 2 − 2 2 − 3 2 + 4 2
3 = − 1 2 + 2 2
we can write the sum for any n by only looking at the possible values of n ( m o d 4 ) .
1. n = 4 k
n = − 1 2 − 2 2 + 3 2 − 4 2 + 5 2 + 6 2 − 7 2 + i = 8 ∑ k + 7 i 2 − ( i + 1 ) 2 − ( i + 2 ) 2 + ( i + 3 ) 2
2. n = 4 k + 1
n = 1 2 + i = 2 ∑ k + 1 i 2 − ( i + 1 ) 2 − ( i + 2 ) 2 + ( i + 3 ) 2
3. n = 4 k + 2
n = − 1 2 − 2 2 − 3 2 + 4 2 + i = 5 ∑ k + 4 i 2 − ( i + 1 ) 2 − ( i + 2 ) 2 + ( i + 3 ) 2
4. n = 4 k + 3
n = − 1 2 + 2 2 + i = 3 ∑ k + 2 i 2 − ( i + 1 ) 2 − ( i + 2 ) 2 + ( i + 3 ) 2
where k denoted any natural number.
Thus the answer is 100.