square number divided by 3 or 4 leave a remainder of 1 or 0, why?

Answer 1

Every third number is a multiple of 3, so for any number n, either n, n-1 or n-2 is a multiple of 3. Take (n-1)^2. If n-1 is a multiple of 3, then obviously (n-1)^2 is too, and thus it would have a remainder of 0 when divided by 3. But notice that (n-1)^2 = (n^2 - 2n +1) = n(n-2) + 1. If (n-1) is not a multiple of 3, then again either n or (n-2) must be a multiple. So n(n-2) must be a multiple of 3. So since n(n-2)+1 is "a multiple of 3, plus 1", then its remainder when divide by 3 is 1. Therefore, every square number divided by 3 gives a remainder of 1 or 0.

Assume the number you're about to square is an even number. Then (2n)^2 = 4n^2. This is a multiple of 4, so dividing by 4 would give a remainder of zero. Now assume it's an odd number. Then it can be written as 2n+1 for some integer n. (2n+1)^2 = (4n^2 + 4n + 1) = 4(n^2 + n) + 1. This is "a multiple of 4 plus 1". So its remainder when dividing my 4 is 1. Therefore, all squares also have a remainder of 0 or 1 when divided by 4.

Answer 2

Because 3 and 4 are only one digit apart. So any square that is divided by 3 or 4 will either go into the square or have a remainder of 1. If the square is divisible by 3, then you automatically know that dividing it by 4 will leave a remainder of 1. The same is true of a square divisible by 4. Dividing the same square by 3 will always have a remainder of 1.

Answer 3

Let the square number be n^2

then we can write
n^2 = 3*p+1 .......(a)
n^2 = 4*p' .......(b) we get

n^2 = 1 / (1- 3p/4p') ----- eq (1)

or n = sqrt [1 / (4p'- 3p)/4p'] ----- eq(2)

p, and p' are integers = 1,2,3,...... so on but no fractions

Now for 'n' can not be imaginary (square root), therefore

4p' - 3p = 1 -eq(3)

and 4p' = has to be a sqaure ----- eq (4)

now give values to p = 1,5,8,21,33.....

from eq (3) and with rider of 4p' being square from e(4)
we get

p' = 1, 4,25/4 (not allowed fraction), 16, 25, .....

these respective sets of p, and p' give us the series of eligible square number (having qualified divisions by 3 & 4, remainder 1)

n^2 = 4,16,64,100, ........ answer

One can vary this if remainder can be varied, which can be buitld in eq (a,b) in place of 1 with different numbers than 3 & 4

EDit: Note for other intermittent chosen values of p the 4p' will not form a square number as assumed in eq (b)

Answer 4

if you divide a number by 3
the possible remainders will be 0, 1, 2

the last digit of a square number can be
0,1,4,5,6,9