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Supposed a function in two variables is defined as $f(a,b) = \frac{a(a-1)b^{2}}{2}$, and $a,b \in N$ where a is fixed on a certain value while b runs from 1 to n.

For instance, $$f(4,1) = \frac{4(4-1)1^{2}}{2} = 6$$ $$f(4,2) = \frac{4(4-1)2^{2}}{2} = 24$$ $$f(4,3) = \frac{4(4-1)3^{2}}{2} = 54$$ $$f(4,4) = \frac{4(4-1)4^{2}}{2} = 96$$ $$f(4,5) = \frac{4(4-1)5^{2}}{2} = 150$$ I noticed that when I take the modulo of this functions in terms of $a$, I get

$$f(4,1) (mod \;4) =2$$ $$f(4,2) (mod \;4) =0$$ $$f(4,3) (mod \;4) =2$$ $$f(4,4) (mod \;4) =0$$ $$f(4,5) (mod \;4) =2$$

But the case is different if $a$ is odd number $$f(3,1) (mod\; 3) = 0$$ $$f(3,2) (mod\; 3) = 0$$ $$f(3,3) (mod\; 3) = 0$$ $$f(3,4) (mod\; 3) = 0$$ $$f(3,5) (mod\; 3) = 0$$

Observation:

  1. If $a$ is odd, the remainder is always 0
  2. If $a$ is even, the remainder is a series of $\frac {a}{2}, 0, \frac {a}{2}, 0, \frac {a}{2}, ...$

Can you give me any idea regarding this?

JERWIN ABLAY
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    In each case, think about which part of your numerator cancels out the 2 in the denominator, and look at the factors of the integer that remains. – Karl Dec 30 '22 at 05:45

1 Answers1

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When $a$ is odd, $\frac{a-1}2$ is an integer, so your fraction is $a\times\frac{a-1}2\times b^2$, which is an integer multiple of $a$, so it's equal to $0$ mod $a$.

When $a$ is even, $\frac a2$ is an integer, so your fraction is $\frac a2\times(a-1)b^2$. This is an integer multiple of $\frac a2$, so mod $a$ it can only be $0$ or $\frac a2$, depending on whether $(a-1)b^2$ is even or odd. Now $a-1$ is odd and $b$ alternates between odd and even as we increment it. As a result, $(a-1)b^2$ also alternates between odd and even, so your remainder alternates between $\frac a2$ and $0$.

Karl
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