Olympiad Problem on Modular Arithmetic Suppose $a,b,c,d$ are integers such that 
$$(3a+5b)(7b+11c)(13c+17d)(19d+23a)=2001^{2001},$$
prove that $a$ is even.
We have $2001=3\cdot 23\cdot 29$, hence we have $3a+5b=3^{e_1}23^{e_2}29^{e_3}$ and similarly for other terms in the product. Then by considering $\pmod 2$, we conclude that $a\equiv c\pmod{2}$ and $b\equiv d\pmod{2}$. Assume on the contrary that $a$ is odd then by considering $\pmod{2}$ again, we have $a,c$ are odd and $b,d$ are even. But then, I couldn't observe anything non-trivial. Please helps.
 A: Look modulo $4$.  You have $(b-a)(-b-c)(c+d)(-d-a)=1\pmod4$.  Assume $b$ and $d$ are even, and $a$ and $c$ are odd, you should be able to find that $a^2c^2=-1\pmod4$
A: Okay so similar to the above answer but I'll add it anyway.
So yeah, you assume $(3a+5b)$ = $3^{e_1}*5^{e_2}*7^{e_3}$, for some $(e_1, e_2, e_3)$ positive integers. We know that the number must be odd, as its prime factorisation does not contain any evens, meaning $(3a+5b)$ mod $2$ is $1$. We say the same about the other three terms. 
Given that the numbers itself are all $1$ mod $2$, and the coefficients multiplied to it in all of the brackets are odd, we know that $a$ mod $2$ $\neq$ $b$ mod $2$, which same can be said for  $c$ mod $2$ $\neq$ $d$ mod $2$. As $2$ only has two possible remainders, we know that:
$a$ mod $2$ = $c$ mod $2$
$b$ mod $2$ = $d$ mod $2$
However the interesting thing comes when you assume $a$ mod $2$ = $1$. Now we see that the left hand side simplifies to $(3)(3)(1)(3) mod 4 = 3^{2001}*21^{2001}*23^{2001} mod 4$ As we can quite easily tell, the RHS is $1$ mod $4$ while the LHS is $3$ mod $4$. This causes a contradiction, leading to the required answer. 
I found this question interesting, and I solved it, so heres my interpretation. 
