Prove that the square of an odd number is always 1 more than a multiple of 4.

Recap, what is 'squaring a number'? Multiplying a number by itself, for example, 3 x 3 = 9. The question hasn't given us a specific number, hence that value can be any number, it is said to be a 'variable'. Normally we assign variables with a letter, it can be any number. Let that number be called 'n'. Any odd number is 1 number greater than an even number. An even number is any number that can be divided by 2 to get a whole number. Therefore, in algebra, let an even number be 2n, so an odd number will be 2n + 1.
If we substitute 'n' for any number into the formula, for instance, 3, 2(3) + 1 = 7, if 7 is squared, we get 49, 1 number greater than 48 (which is a multiple of 4). If we substitute 5 for 'n', we get 2(5) + 1 = 11, 11 squared is 121, which is 1 number greater than 120 (which is a multiple of 4). Hence, we have proved that the square of an odd number is always 1 more than a multiple of 4.