75th NMO Selection Tests

We will prove that there are only two families of solutions: $$(1,1,2^x-1),(1,2^x-1,2^x+1),$$ where $x$ is a positive integer.

Step I: $a,b,c$ are odd and pairwise coprime in pairs. It is easy to see that $$a\equiv b\equiv c(\mathrm{mod}2).$$ If $a,b,c$ are all even, denote by $d,e,f$ the exponents of 2 in their prime factorization, respectively, with $1\le d\le e\le f$. Then the exponent of 2 in $bc+a$ is $d$. Since $bc+a$ is a power of 2, we get $$bc+a=2^d,$$ which is impossible because $2^d\le a$. Thus, $a,b,c$ must be odd numbers. If $a,b$ have a common odd prime factor $p$, then $p$ divides $ac+b$, so $ac+b$ is divisible by $p$ and thus cannot be a power of 2. Similarly, one shows $(a,c)=1$ and $(b,c)=1$, which completes Step I.

Hence, we can assume $a\le b\le c$, where $a,b,c$ are odd positive integers pairwise coprime such that $$ab+c,ac+b,bc+a$$ are powers of 2. Since $a\le b\le c$, we have $$ab+c\le ac+b\le bc+a.$$ Step II: $a=1$. Suppose, by contradiction, that $1<a$. Then $1<a<b<c$, and $$ab+c=2^k,ac+b=2^m,bc+a=2^n,$$ with $3\le k<m<n$ (since $2^k\ge 3+5=8$). From the above equalities, we get the congruences: $$ab\equiv -c(\mathrm{mod}{2}^k),ac\equiv -b(\mathrm{mod}{2}^k),bc\equiv -a(\mathrm{mod}{2}^k).$$ Multiplying these congruences and noting that $a,b,c$ are odd, we obtain $$abc\equiv -1(\mathrm{mod}{2}^k)$$ and $$a^2\equiv b^2\equiv c^2\equiv 1(\mathrm{mod}{2}^k)$$ (by multiplying the first congruence by $c$, the second by $b$, and the third by $a$).

Since for any odd natural number $x$, $$\gcd (x-1,x+1)=2,$$ we deduce that $$a\equiv \pm 1(\mathrm{mod}{2}^{k-1}),b\equiv \pm 1(\mathrm{mod}{2}^{k-1}),c\equiv \pm 1(\mathrm{mod}{2}^{k-1}).$$ From these congruences, it follows that $$a,b,c\ge 2^{k-1}-1.$$ This inequality implies $$2^k=ab+c\ge (2^{k-1}-1{)}^2+2^{k-1}-1=2^{2k-2}-2^{k-1},$$ which means $2\ge 2^{k-1}-1\ge 3$, a contradiction since $k>3$. Therefore, the assumption is false and $a=1$.

Step III: the solutions. If $b=1$, then clearly $c=2^x-1$. Otherwise, assume $a=1$ and $1<b<c$. From Step II, we know that $c>b\ge 2^{k-1}-1$. Since $b+c=2^k$ and $b,c$ are odd, the only possibility is $b=2^{k-1}-1$, $c=2^{k-1}+1$. Hence, the only solutions are: $(1,1,2^x-1)$ and $(1,2^x-1,2^x+1)$, $x\in {\mathbb{N}}^{\ast }$.