Pass the Odds

Adding both the equations we get

$a + 2b + 2c + d = odd + odd = even$

Since $2b + 2c$ will be even

Therefore $a + d = even -(2b+2c) = even - even = even$

$a+b+c$ is odd.

$b+c+d$ is also odd.

Subtracting we get $a-d$ = odd-odd = even.

Since $a-d$ is even $a+d$ will also be even.

Since the other numbers in the 2 sums are the same, then a and d must have the same parity. Either both are odd or both even, but a+d will be even.

Adding both the equations we get

a+2b+2c+d = odd+odd = even number

Since 2b+2c will be even

Therefore a+d = even number-(2b+2c) = even

Let b+c be odd. a+b+c is odd so a is even, similarly d is even. a+d is even

Let b+c be even a+b+c is odd so a is odd, similarly d is odd. a+d is even.

The problem can be slightly generalised to -

If $a,b,c$ and $d$ are integers such that $a+b+c$ and $b+c+d$ have the same parity, is $a+d$ odd or even?

By 'have the same parity' I mean 'are either both even or both odd', or in more technical language 'are the same modulo 2'.

The answer to this more general problem is $\boxed{a+d\text{ is even}}$

Proof (all working is modulo 2)

Given

$a+b+c=b+c+d$

Subtract $b+c$ from both sides to get

$a=d$

and so

$a+d=a+a=2a=0$

and so

$\boxed{a+d\text{ is even}}$