Question 37
Prove that if $a, b, c$ and $d$ are non-negative integers, and it is
$$
(a+b)^2+2a+b=(c+d)^2+2c+d
$$
It must necessarily be $a=c$ and $b=d$.
Prove the same conclusion if, instead of $(*)$, it holds
$$
(a+b)^2+3a+b=(c+d)^2+3c+d
$$
See that, on the other hand, there are non-negative integers $a \neq c, b \neq d$, such that
$$
(a+b)^2+4a+b=(c+d)^2+4c+d
$$
$$
(a+b)^2+2a+b=(c+d)^2+2c+d
$$
It must necessarily be $a=c$ and $b=d$.
Prove the same conclusion if, instead of $(*)$, it holds
$$
(a+b)^2+3a+b=(c+d)^2+3c+d
$$
See that, on the other hand, there are non-negative integers $a \neq c, b \neq d$, such that
$$
(a+b)^2+4a+b=(c+d)^2+4c+d
$$