National Olympiad of Argentina

Assume $a\ge b$ by symmetry; then $0<c<b\le a<99$. Also $2a^2\ge a^2+b^2>99^2$, so that $a\ge 71$. Thus $71\le a\le 98$. Write the equation as $(b+c)(b-c)=(99-a)(99+a)$.

For $\min (a+b+c)$ look at $a=98$ and $a=97$ first. If $a=98$ then $(b+c)(b-c)=197$, and $197$ is a prime. Hence $b+c=197$, $a+b+c=295$. If $a=97$ then $(b+c)(b-c)=2\cdot 196=392=14\cdot 28$, and $392$ has no representation $392=d_1{d}_2$ with $d_1,d_2$ between $14$ and $28$. The sum of integers with a fixed product is a minimum when the factors are as close as possible, therefore $b+c\ge 28$ and $a+b+c\ge 97+28=125$. The equality $a+b+c=125$ is attained for $b+c=28$, $b-c=14$, i.e. $b=21$, $c=7$; the triple $(97,21,7)$ is admissible. We show that $125$ is the desired minimum.

Note that $(b+c)(b-c)=(99-a)(99+a)$ implies $b+c>\sqrt{99^2-a^2}$, hence $a+b+c>a+\sqrt{99^2-a^2}$. So a sufficient condition for $a+b+c>125$ is $a+\sqrt{99^2-a^2}>125$. This is equivalent to $a^2-125a+2912<0$. The quadratic function $f(t)=t^2-125t+2912$ increases in $[62\frac{1}{2},+\infty )$ and $f(94)<0$, implying $a+b+c>125$ for $a\in [71,94]$. If $a=95$ then $(b+c)(b-c)=4\cdot 194$. As $b+c>b-c$ and $b+c$, $b-c$ have the same parity, we see that $b+c\ge 194>125$. If $a=96$ then $(b+c)(b-c)=3\cdot 195=15\cdot 39$, and the factors in the last product are closest possible. Hence $b+c\ge 39$ and $a+b+c\ge 96+39=135>125$.

For the maximum of $a+b+c$ note first that $a+b+c$ is odd. Also $99-a<b-c<b+c<99+a$ in view of $0<c<b\le a<99$ and $(b+c)(b-c)=(99-a)(99+a)$. Moreover the four numbers $b+c$, $b-c$, $99-a$, $99+a$ have the same parity, so $b-c=(99-a)+2k$ with $k\ge 1$ an integer. We prove that the maximum is attained when $k=1$, i.e. $b-c=101-a$.

First restrict attention to admissible triples $(a,b,c)$ that satisfy the last additional condition. Set $x=b-c=101-a$ for clarity. Then $a=101-x$, and one can express $a+b+c$ in terms of $x$: $$a+b+c=a+\frac{(99-a)(99+a)}{b-c}=303-2\left(x+\frac{200}{x}\right).$$ Since $a+b+c$ is odd, $x+\frac{200}{x}$ is an integer. So $x$ and $\frac{200}{x}$ are divisors of $200$ with product $200$. We have to minimize their sum in order that $a+b+c$ be a maximum. A factorization $d_1{d}_2=200$ is needed with $d_1$ and $d_2$ closest possible, which leads to $\{x,\frac{200}{x}\}=\{10,20\}$. Therefore $a+b+c\le 303-2(10+20)=243$. The value $243$ is attained for the triple $(91,81,71)$ which is admissible.

If $b-c\ne 101-a$ then $b-c\ge (99-a)+4=103-a>0$, hence $$a+b+c=a+\frac{(99-a)(99+a)}{b-c}\le a+\frac{(99-a)(99+a)}{103-a}.$$ To complete the proof it suffices to show that $a+\frac{(99-a)(99+a)}{103-a}<243$ whenever $a<99$. Use the last inequality to reach the equivalent form $a^2-73a+7614>0$. The latter holds for all real $a$ since the trinomial $a^2-73a+7614$ has a negative discriminant.