China Southeastern Mathematical Olympiad

From condition (2), we get $$(a+1)(c+1)=(b+1{)}^2.$$ Set $a+1=n^{2}x$, $c+1=m^{2}y$, with no square factor larger than 1 in $x$, $y$, then we could get that $x=y$. This is due to the fact that from above, we have $$(mn{)}^{2}xy=(b+1{)}^2,$$ which means $mn\mid (b+1)$. Set $b+1=mn\cdot w$, then the above can be simplified to $$xy=w^2.$$ If $w>1$, then the prime number $p_1\mid w\Rightarrow p_1^2\mid w^2$. As there is no square factor larger than 1 in $x$, $y$, then $p_1\mid x$ and $p_1\mid y$. Now set $x=p_1{x}_1$, $y=p_1{y}_1$, $w=p_1{w}_1$. Then the above can be simplified to $$x_1{y}_1=w_1^2.$$ If $w_1>1$ still exists, then there will be a new prime number $p_2\mid w_1\Rightarrow p_2^2\mid w_1^2$. As there is no square factor larger than 1 in $x_1$, $y_1$, then $p_2\mid x_1$ and $p_2\mid y_1$. Now set $$x_1=p_2{x}_2,y_1=p_2{y}_2,w_1=p_2{w}_2.$$ Then the above can be simplified to $x_2{y}_2=w_2^2,\cdots$. Since there are finite prime factors of $w$, then carrying on as above, we obtain that there exists $r$, such that $w_r=1$. $x_r{y}_r=w_r^2\Rightarrow x_r=y_r=1$, we have $x=p_1{p}_2\cdots p_r=y$ as desired. Now we can set $x=y=k$, and have $$\{\begin{array}{l}a=k{n}^2-1,\\ b=kmn-1,\\ c=k{m}^2-1,\end{array}$$ where $$1\le n<m,a<b<c<100,$$ with no square factor larger than 1 in $k$ and $k\ne 1$. Otherwise, if $k=1$, then $c=m^2-1$. As $c$ is larger than the third prime number 5, thus $c=m^2-1>5\Rightarrow m\ge 3$ and $$c=m^2-1=(m-1)(m+1)$$ is a composite number. Contradiction! Hence, $k$ is either a prime number, or the product of several different prime numbers (that is, $k$ is larger than 1 and with no square factor larger than 1 in $k$). We say that "$k$ has the property $p$".

a. From above, $m\ge 2$. When $m=2$, then $n=1$ and $$\{\begin{array}{l}a=k-1,\\ b=2k-1,\\ c=4k-1.\end{array}$$ Since $c<100\Rightarrow k<25$, then if $k\equiv 1(\mathrm{mod}3)$, we get $3\mid c$ and $c>3$, which means $c$ is a composite number. If $k\equiv 2(\mathrm{mod}3)$, then when it is even, the $k$ satisfying the property $p$ is 2 or 14, where the corresponding $a=2-1=1$ and $b=2\cdot 14-1=27$ are not prime numbers. On the other hand, when it is odd, the $k$ satisfying the property $p$ is 5, 11, 17 or 23, where all the corresponding $a=k-1$ are not prime numbers. If $k\equiv 0(\mathrm{mod}3)$, the $k$ satisfying the property $p$ is 3, 6, 15 or 21. When $k=3$, we get the first solution $$f_1=(a,b,c)=(2,5,11).$$ When $k=6$, the second solution is $$f_2=(a,b,c)=(5,11,23).$$ But when $k=15,21$, the corresponding $a=k-1$ are not prime numbers.

b. When $m=3$, then $n=2$ or 1. If $m=3,n=2$, we have $$\{\begin{array}{l}a=4k-1,\\ b=6k-1,\\ c=9k-1.\end{array}$$ Since $c\le 97\Rightarrow k\le 10$, then the $k$ satisfying the property $p$ is 2, 3, 5, 6, 7 or 10. When $k=3,5,7$, the corresponding $c=9k-1$ are all composite numbers. When $k=6,b=6k-1=35$, which is a composite number. When $k=10,a=4k-1=39$, which is also a composite number. But when $k=2$, we get the third solution $$f_3=(a,b,c)=(7,11,17).$$ If $m=3,n=1$, we have $$\{\begin{array}{l}a=k-1,\\ b=3k-1,\\ c=9k-1.\end{array}$$ As $k\le 10$, the $k$ satisfying the property $p$ is 2, 3, 5, 6, 7 or 10. When $k=3,5,7,$ the corresponding $b=3k-1$ are all composite numbers. When $k=2,10$, the corresponding $a=k-1$ are not prime numbers. But when $k=6$, we get the fourth solution $$f_4=(a,b,c)=(5,17,53).$$

c. When $m=4$, from $c=16k-1\le 97\Rightarrow k\le 6$, then the $k$ satisfying the property $p$ is 2, 3, 5 or 6. When $k=6$, $c=16\cdot 6-1=95$, which is a composite number. When $k=5$, then $$\{\begin{array}{l}a=5n^2-1,\\ b=20n-1.\end{array}$$ As $n<m=4$, $n$ can be 1, 2, 3, which means at least one of $a,b$ is not a prime number. When $k=3,c=48-1=47$ and $$\{\begin{array}{l}a=3n^2-1,\\ b=12n-1.\end{array}$$ Considering $n<m=4$, the corresponding $a,b$ are both composite numbers if $n=3$. But, if $n=2$, we get the fifth solution $$f_5=(a,b,c)=(11,23,47).$$ If $n=1$, we get the sixth solution $$f_6=(a,b,c)=(2,11,47).$$ When $k=2$, $c=16\cdot k-1=31$ and $$\{\begin{array}{l}a=2n^2-1,\\ b=8n-1.\end{array}$$ Since $n<m=4$, the seventh solution $$f_7=(a,b,c)=(17,23,31)$$ exists if $n=3$.

d. When $m=5$, then $c=25k-1\le 97$ and the $k$ satisfying the property $p$ is 2 or 3, but the corresponding $c=25k-1$ are both composite numbers.

e. When $m=6$, then $c=36k-1\le 97$ and the $k$ satisfying the property $p$ is 2. Hence, $c=2\cdot 36-1=71$ and $$\{\begin{array}{l}a=2n^2-1,\\ b=12n-1.\end{array}$$ As $n<m=6$, the eighth solution $$f_8=(a,b,c)=(7,23,71)$$ exists if $n=2$, and the ninth solution $$f_9=(a,b,c)=(31,47,71)$$ exists if $n=4$.

f. When $m=7$, then $c=49k-1\le 97$ and the $k$ satisfying the property $p$ is 2. Hence, $c=2\cdot 49-1=97$ and $$\{\begin{array}{l}a=2n^2-1,\\ b=14n-1.\end{array}$$ As $n<m=7$, the tenth solution $$f_{10}=(a,b,c)=(17,41,97)$$ exists if $n=3$, and the eleventh solution $$f_{11}=(a,b,c)=(71,83,97)$$ exists if $n=6$.

g. When $m\ge 8$, then $c=64k-1\le 97$, but the $k$ satisfying the property $p$ does not exist.

Therefore, there are 11 possible solutions, namely $f_1,f_2,\dots ,f_{11}$.