No, there cannot be any.
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If n is an odd integer then the next two consecutive odd integers are n+2 and n+4.
If, for an n*n matrix, A, there exists a matrix B such that AB = I, where I is the n*n identity matrix, then the matrix B is said to be the inverse of A. In that case, BA = I (in general, with matrices, AB ≠BA) I is an n*n matrix consisting of 1 on the principal diagonal and 0s elsewhere.
A matrix is an array of elements in m rows and n columns.
Every integer is either even (divisible by 2) or odd (not divisible by 2). Since an even number plus 1 is odd and an odd number plus one is even, because 1 does not divide 2. We know (n + 4) is odd. The next integer is (n + 4 + 1) = (n + 5), because an odd number plus 1 is even, (n + 5) is even. The integer after (n + 5) is (n + 6), since (n + 5) we know is even, (n + 6) must be odd. Since (n + 6) is the smallest integer that is greater than (n + 4) and is odd, so (n + 6) is the next odd integer.
No even values of n will give give an odd value of n3, so n must be odd.When n is odd, n2 is also odd, so n2+1 must be even. ■