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Squares of odd numbers are odd, and are congruent to 1 modulo 8, since (2n + 1) 2 = 4n(n + 1) + 1, and n(n + 1) is always even. In other words, all odd square numbers have a remainder of 1 when divided by 8. Every odd perfect square is a centered octagonal number. The difference between any two odd perfect squares is a multiple of 8.
[7] [8] [9] It is widely believed, [10] but not proven, that no odd perfect numbers exist; numerous restrictive conditions have been proven, [10] including a lower bound of 10 1500. [11] The following is a list of all 52 currently known (as of January 2025) Mersenne primes and corresponding perfect numbers, along with their exponents p.
The number of divisors of a perfect number (whether even or odd) must be even, because N cannot be a perfect square. [51] From these two results it follows that every perfect number is an Ore's harmonic number.
A ramification of the difference of consecutive squares, Galileo's law of odd numbers states that the distance covered by an object falling without resistance in uniform gravity in successive equal time intervals is linearly proportional to the odd numbers. That is, if a body falling from rest covers a certain distance during an arbitrary time ...
All odd squares are ≡ 1 (mod 8) and thus also ≡ 1 (mod 4). If a is an odd number and m = 8, 16, or some higher power of 2, then a is a residue modulo m if and only if a ≡ 1 (mod 8). [7] For example, mod (32) the odd squares are 1 2 ≡ 15 2 ≡ 1 3 2 ≡ 13 2 ≡ 9 5 2 ≡ 11 2 ≡ 25 7 2 ≡ 9 2 ≡ 49 ≡ 17. and the even ones are 0 2 ...
Because m is powerful, each prime factor with an odd exponent has an exponent that is at least 3, so m/b 3 is an integer. In addition, each prime factor of m/b 3 has an even exponent, so m/b 3 is a perfect square, so call this a 2; then m = a 2 b 3. For example: = =,
Legendre's conjecture: Does there always exist at least one prime between consecutive perfect squares? Are there infinitely many primes p such that p − 1 is a perfect square? In other words: Are there infinitely many primes of the form n 2 + 1? As of 2025, all four problems are unresolved.
A most-perfect magic square of order n is a magic square containing the numbers 1 to n 2 with two additional properties: Each 2 × 2 subsquare sums to 2 s , where s = n 2 + 1. All pairs of integers distant n /2 along a (major) diagonal sum to s .