Let's say we want our integer number pair $(x,y)$ that ranges from $2$ to $99$ to satisfy $x+y = a$ and $xy=b$. My question is, is there $a$ and $b$, integers that have more than one pair $(x,y)$ that satisfies $x+y = a$ and $xy=b$?
If so, what is the condition imposed to $a,b$ that makes it have more than one pair?
Because $(X-x)(X-y)=X^2-(x+y)X+xy$, the numbers $x,y$ are the two solutions of $$X^2-aX+b=0 $$ Apart from switching their order, there is no choice, they are $\frac{a\pm\sqrt{a^2-4b}}{2}$. Thus no two pairs can exist, no matter what $a,b$ are.
This boils down to the uniqueness of the set of roots of a (quadratic) polynomial (which holds true in a commutative ring $R\iff R$ is an integral domain, i.e. $\,xy = 0\Rightarrow x=0\,$ or $\,y=0)$
Lemma $\ $ TFAE for $\,A,B,a,b\in \Bbb Q\,$ (or $\Bbb R$ or $\Bbb C$ or any integral domain)
$(1)\quad A+B = a+b,\,\ AB = ab,\ $ i.e. $\,\{A,B\}\ \&\ \{a,b\}\,$ have equal sum & product
$(2)\quad (X\!-\!A)(X\!-\!B) = (X\!-\!a)(X\!-\!b),\ $ i.e. equal polynomials in $X$, i.e. equal coef's
$(3)\quad \{A,B\} = \{a,b\},\ \ $ i.e, equal multi-sets
Proof $\ (1\!\Rightarrow\! 2)\ $ Expanding both polynomials shows they have equal coef's, after applying $(1)$.
$(2\!\Rightarrow\! 3)\ $ Eval $(2)$ at $X = A\Rightarrow (A-a)(A-b) = 0\,$ so, since $D$ is a domain, $A = a\,$ or $A = b.\,$ Wlog $A = a,\,$ so eval $(2)$ at $\,X = a+1\Rightarrow X-B = X-b,\,$ so $\,B = b$.
$(3\!\Rightarrow\! 1)\ $ Equal multi-sets have equal sum & product (since both operations are commutative)
Remark $ $ Generally if fails in non-domains (rings with zero-divisors) since if $\,ab=0,\,\ a,b\ne 0\,$ then $\, (X\!-\!a)(X\!-\!b) = X(X\!-\!a\!-\!b)$ and $\{a,b\}\ne \{0,a+b\}.\,$ So the equalities in the Lemma are equivalent for a ring iff the ring is an integral domain, i.e. it has no zero-divisors, i.e. all elements $\ne 0$ are cancellable. Another way to view it is that $D$ is a domain iff $\,X\!-\!a\,$ is prime in $\,D[X],\,$ and the the uniqueness of the root sets corresponds to the uniqueness of the associated prime factorization $\,(X\!-\!a)(X\!-\!b),\,$ see here. More generally we can use the uniqueness of linear factorizations $f = (X-a_1)\cdots (X-a_n)\,g,\,$ where $g$ has not roots.
Suppose $x+y=x'+y'$ with $x\geq y$ and $x'\geq y'$.
Let $M=(x+y)/2=(x'+y')/2.$ Let $D=x-M=M-y$ and $D'=x'-M=M=y'.$ We have $D\geq 0$ and $D'\geq 0$.
Then $xy=(M+D)(M-D)=M^2-D^2$ and $x'y'=(M+D')(M-D')=M^2-D'^2.$
So $xy=x'y'\implies D^2=D'^2\implies D=D'$ (because $D,D'$ are non-negative) . But then $x=M+D=M+D'=x',$ and $y=M-D=M-D'=y'.$
Corollary: Two rectangles with the same perimeter and the same area are congruent.
It depends whether your $(x,y)$ is an ordered pair or not.
If $b=\frac {a^2}4$ then there is only one ordered pair. Otherwise if there is one ordered pair, there are two.
