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**Deveno** There is "something missing" from your proof....can you see what it is? I'll give you a hint: what would happen if you tried to use $\displaystyle \mathbb{Z}[x,y]$ instead?

The most common way of proving two sets are equal in mathematics is to show they contain each other. Clearly, <x+y, x-y> is contained in the ideal generated by x and y, as both x+y and x-y are in <x,y> and <x+y,x-y> is the MINIMAL ideal containing x+y and x-y.

So the "hard" part is showing that <x,y> is contained in <x+y, x-y>, which amounts to showing that both x and y are elements of <x+y,x-y>. And:

x = (1/2)(x+y) + (1/2)(x-y) (note that we need to use rational numbers, integers wouldn't do!)

y = (1/2)(x+y) - (1/2)(x-y), both of which are clearly elements of <x+y, x-y>.