Hello, kohila!
I have a primitive solution with brute-force Listing.
Let
Since is even, either (1) all are even or (2) exactly one is even.
Hence, there are four cases to consider:
. . (1) are even.
. . (2) even, odd.
. . (3) even, odd.
. . (4) even, odd.
. . .Since is odd, must be odd.
. . .Hence: is even, is odd.
. . .Since is odd, must be odd.
. . .Hence, is even, is odd.
. . .Since is odd, must be even.
. . .Hence, is odd, is even.
. . .Since is odd, must be even.
. . .Hence, is odd, is even.
In all cases, and have opposite parity.
Hence, their squares will have opposite parity.
The difference between two numbers of opposite parity is odd.