Proof that $a^2 + b^2 + ab$ and $a^2 + b^2 - ab$ can not both be perfect squares If $a$, $b$ are co-prime integers, how can I prove that $a^2 + b^2 + ab$ and $a^2 + b^2 - ab$ cannot be perfect squares? I know that perfect squares should be capable of expression in the form $a^2 + b^2 + 2ab$ and $a^2 + b^2 - 2ab$, but it is not immediately clear to me that there might not be other integers $x$, $y$ that would work even if $a$, $b$ do not. Thanks if anybody can help.

I should have said that $a$, $b$ are distinct integers and are both positive so that they could not both be equal to 1.

Please note that the premise of this question has now been disproved with a straightforward counter example, which is all the answer I require. Thanks to everyone who has taken the trouble to contribute. I am grateful for the help but won't be continuing to monitor further responses.
Above simultaneous equations shown below:

$(a^2+ab+b^2)=p^2$

$(a^2-ab+b^2)=q^2$

I don't think there are integer solution's to above equation. 

But if "OP' is interested there is solution in irrational's 

and is given below:

$(p,q)=((m^2+mn+n^2),(m^2-mn+n^2))$

where,  $(m,n)$=$((2+5^{1/2}),(1))$

$(a,b))$=$((8+4*5^{1/2}),(5+2*5^{1/2}))$

$(p,q)$=$((12+5*5^{1/2}),(8+3*5^{1/2}))$
Above simultaneous equations shown below:

$(a^2+ab+b^2)=p^2$

$(a^2-ab+b^2)=q^2$

I don't think there are integer solution's to above equation. 

But if "OP' is interested there is solution in irrational's 

and is given below:

$(p,q)=((m^2+mn+n^2),(m^2-mn+n^2))$

where,  $(m,n)$=$((2+5^{1/2}),(1))$

$(a,b))$=$((8+4*5^{1/2}),(5+2*5^{1/2}))$

$(p,q)$=$((12+5*5^{1/2}),(8+3*5^{1/2}))$