is A an even number?

Question

Let $a,b,c,d$ be positive integers such that

$(3a+5b)(7b+11c)(13c+17d)(19d+23a)=2001^{2001}$

hence, prove that $a$ is even.

I tried to approach this problem reducting it modulo 6. From which we get $(3a-b)(b-c)(c-d)(d-a)\equiv3$ $\mod 6$ The combinations such that $XYZT\equiv3$ $\mod 6$ are only

$1,1,1,3$;

$1,1,3,3$;

$1,3,3,3$;

$3,3,3,3$;

Wich only gets $a\equiv2,4$ $\mod 6$ hence $a$ is even.

Another way to approach this problem was to sum all the factors, wich is 0 mod 2. But is only 0 mod 4 if a were even. But I didn't get very far.

First I would like to know if my solution is correct. And if someone has a better solution please share with us. Thanks.

Answer 1

The original question asks to prove the statement by observing the expression with $\mod 4$.

We have $(-a+b)(-b-c)(c+d)(-d-a)=1 \mod 4$.

We find that $(ac)^2\equiv -1 \mod 4$

If we assume that $a,c$ are odd and $b,d$ are even, then we see that this is not possible since when we have an odd number $a\equiv 1 \mod 4,\quad a^2\equiv 1 \mod 4$ and when $a\equiv -1 \mod 4, \quad a^2\equiv 1\mod 4$.

Thus, $a$ must be even.

Answer 2

i didn't understand the accepted answer, so i posted this.

modulo 4, the sum of the four factors is 2a, and their product is 1. the four factors must be {1,1,1,1},{3,3,3,3}, or {1,1,3,3}. in all three cases, their sum is 0.
$2a\equiv0\pmod4 \iff a$ is even