Show that there are exactly $2$ numbers $a$ in the set $\{1,2,3\dots9400\}$ such that $a^2-a$ is divisible by $10000$
Use Modular arithmetic and concepts of coprime numbers
we know
$10000=2^4*5^4$
In order for $10000$ to divide $a^2-a$ both $2^4$ and $5^4$ must divide $ a^2-a $
Write $a^2-a=a(a-1)$
Note that $a$ and $a-1$ are coprime numbers so they can not have any common divisors
That means either
$2^4$ divides $a$ and $5^4$ divides $(a-1)$
or
$2^4$ divides $a-1$ and $5^4$ divides $a$
$2^4$ divides $a$ and $5^4$ divides $(a-1)$
AS $5^4=625$ divides $a-1$ we can write
$a-1=625k$ for some integer $k$
$\Rightarrow a=625k+1$
Now $2^4=16$ divides $a=625k+1$
Note that $16$ divides $624$ hence $16$ will also divide $624k$
We can write $625k+1=624k+k+1$
As $16$ divides $625k+1=624k+k+1$ so it must divide $k+1$
$\Rightarrow k$ is among the numbers $T=\{15,31,47\dots\}$
For $k=15$ we get $a=625*15+1=9376$
For $k=31$ the value of $a$ is $19376>9400$
So from $31$ onward none of the elements of $T$ is acceptable
So in this case the only possible value of $a$ is $9376$
$2^4$ divides $a-1$ and $5^4$ divides $a$
As $5^4=625$ divides $a$ so $a=625k$ for some integer $k$
$\Rightarrow a-1=625k-1$
Now $2^4=16$ divides $a-1$
Write $625k-1=624k+k-1$
As in the last case we can argue that $16$ divides $624$ so it must divide $k-1$
So $K$ must be from the set $U=\{1,17,33\dots\}$
For $k=17 , a=10625>9400$ so any value of $k$ greater or equal $17$ does not count
So the only possible value that $k$ can take in this case is $k=1$
So $a=625$
So $625$ & $9376$ these are the only $2$ values that $a$ can take
