User:RandomUndergrad

Lacklustre Bio
Currently a final year Math student at CUHK.

Primes Expressible as $x^2+ny^2$
I'll put this here for (my) future reference:

int main{int n,x,y;for(n=1;n<11;n++)for(x=1;x<33;x++)for(y=1;y<33;y++)if(x*x+y*y*n==1009)printf("1009=%d^2+%dx%d^2\n",x,n,y);}
 * 1) include 

also


 * 1129=20^2+1x27^2
 * 1129=27^2+1x20^2
 * 1129=29^2+2x12^2
 * 1129=19^2+3x16^2
 * 1129=27^2+4x10^2
 * 1129=2^2+5x15^2
 * 1129=23^2+6x10^2
 * 1129=11^2+7x12^2
 * 1129=29^2+8x6^2
 * 1129=20^2+9x9^2
 * 1129=33^2+10x2^2

Sum of Product of $k$ Consecutive Integers
In particular, when $m = 1$, the theorem takes a nice form:

For $l = 0$, $\displaystyle \sum_{i = 1} ^ n i = \dfrac {n (n + 1)} 2$ which is Closed Form for Triangular Numbers

For $l = 1$, $\displaystyle \sum_{i = 1} ^ n i (i + 1) = \dfrac {n (n + 1) (n + 2)} 3$ which is Sum of Sequence of Products of Consecutive Integers

Credit: The relationship between binomial coefficients and products of consecutive integers was noticed by my friend Oscar, whose observation allowed for a intuitive view to the validity of the equation.

My original proof for the cases $m = 0$ was an induction on $n$ with a fixed $l$.

For $n = 1$, $RHS = \dfrac {1 (2) \cdots (1 + l + 1)} {l + 2} = (l + 1) ! = LHS$

For the induction step, assume $\displaystyle \sum_{i = 1} ^ m i (i + 1) \cdots (i + l) = \dfrac {m (m + 1) \cdots (m + l + 1)} {l + 2}$.

Then

The general theorem results from subtraction.

I wish to generalize this result to arithmetic progressions at some point.