Show if a and b are congruent to x mod 4 and y mod 4...

Question

Yes this is a homework assignment. It has me quite lost on where to go ahead though.

Let $a,b,x,y \in \mathbb{Z}$ . Show that: if $a\equiv$₄ $x$ and $b\equiv$₄ $y$, then $3a^2 +5b - 7\equiv 3x^2 +5y - 7$.

It is for an Abstract Algebra course, and I cannot use anything more complex than the basic Algebraic Axioms.

I know $4|(b-y)$ and $4|(a-x)$, besides that I am stumped.

Thanks for any thoughts or help!

Edit:
Solution uses:

There are $s,t \in \mathbb{Z}$ such that $a-x=4s$ and $b-y=4t$.

$(3a^2 +5b-7)-(3x^2 +5y-7)=3a^2+5b-7-3x^2-5y+7=3(a^2-3x^2)+5(b-y) =3(a+x)(a-x)+5(b-y)=3(a+x)(4s)+5(4t)=4[3(a+x)+5]$

Since $3(a+x)+5 \in \mathbb{Z}$, ...

The rest gets to the solution.

Answer 1

Guide:

First compute

$$3x^2+5y-7 - (3a^2+5b-7 )$$

Try to have terms like $(b-y)$ and $(x-a)$, remember that these terms are divisible by $4$.

Useful identity:

$$p^2-q^2=(p-q)(p+q)$$

Answer 2

Hint:   prove first that if $p\equiv_n r$ and $q\equiv_n s$ then $p\cdot q\equiv_n r \cdot s\,$ and $p+ q\equiv_n r + s\,$, in other words congruences can be added and multiplied together just like equalities.

It then follows that $a^2\equiv_4 x^2\,$, therefore $3a^2\equiv_4 3x^2\,$, and so $\;\ldots$