Let $\gcd(b,c)=d$, $b=d.x$ and $c=d.y$, where $\gcd(x,y)=1$ We have $\gcd(a,dx)+\gcd(a,dy)+d=dx+dy+2023$ Let $\gcd(a, d)=s$ $\implies$ $a=s.t, d=s.q$ and $\gcd(t, q)=1$. $s|\gcd(a, dx)$, $s|\gcd(a, dy)$, $s|dx, s|dy$, $s|d$ $\implies$ $s|2023$ $\gcd(a, dx)=\gcd(st, sqx)=s.\gcd(t, qx)=s.\gcd(t, x)$ Similarly, $\gcd(a, dy)=s.\gcd(t, y)$ Let $2023=k.s$ We have $s.\gcd(t, x)+s.\gcd(t, y)+s.q=s.q.x+s.q.y+k.s$ $\iff$ $\gcd(t, x)+\gcd(t, y)+q=q.x+q.y+k$. $q=q.x-\gcd(t,x)+q.y-\gcd(t,y)+k$ $\geq q.x-x+q.y-y+k=(x+y)(q-1)+k$ $\iff q-1\geq (x+y)(q-1)+k-1$ $\implies q-1\geq (x+y)(q-1) \implies q=1\implies s=d\implies 0\geq k-1\implies k=1\implies s=2023\implies d=2023$. Hence, $\gcd(b, c)=2023$

I think this is harder than senior P1, but oh well. Notice that if $b \mid a$ and $c \mid a$, then $\gcd(a,b)=b$ and $\gcd(a,c)=c$, so we get that $\gcd(b,c)=2023$ as needed. Suppose that $b \nmid a$ and $c \nmid a$, then $\gcd(a,b) \le \frac{b}{2}$ and $\gcd(a,c) \le \frac{c}{2}$, because $\gcd(a,b) = \gcd(a \pmod {c}, c) \le \frac{c}{2}$ . So we get that \begin{align*} b+c+2023=\gcd(a,b)+\gcd(a,c)+\gcd(b,c) \le \frac{b}{2}+\frac{c}{2}+\gcd(b,c) \\ b+c+4046 \le 2\gcd(b,c) =\gcd(b,c)+\gcd(b,c) \le b+c \end{align*}The last inequality is a contradiction. Suppose. that $b \mid a$, but $c \nmid a$. Then $c \nmid b$, because otherwise if it would be the case we have $b \mid a$, $c \mid b$ implies $c \mid a$, which contradicts our assumption. Notice $\gcd(a,b)=b$, $\gcd(a,c) \le \frac{c}{2}$ and $\gcd(b,c) \le \frac{c}{2}$. Therefore \begin{align*} b+c+2023 =gcd(a,b)+\gcd(a,c)+\gcd(b,c) \\ c+2023 = \gcd(a,c)+\gcd(b,c) \le \frac{c}{2}+\frac{c}{2} = c \end{align*}The last inequality is a contradiction. Similarly, we can show that $b \nmid a$, $c \mid a$ gives us a contradiction. This exhausts all the possible cases.

Let $\gcd(a,b)=d \implies d \mid a$ , $d \mid b \implies d \leq a$ , $d \leq b$ $\gcd(a,c)=k \implies k \mid a$ , $k \mid c \implies k \leq a$ , $k \leq c$ $\gcd(b,c)=l \implies l \mid b$ , $l \mid c \implies l \leq b$ , $l \leq c$ From $d \leq b$ and $k \leq c$ we get: $d+k \leq b+c$ $...(*)$ Let's suppose that $gcd(b,c) < 2023 \implies l<2023$ From the condition we have: $\gcd(a,b)+\gcd(a,c)+\gcd(b,c)=b+c+2023 \implies d+k+l=b+c+2023.$ Combining with $l<2023$ we have: $d+k+l<d+k+2023 \implies b+c+2023=d+k+l<d+k+2023 \implies b+c+2023<d+k+2023 \implies b+c<d+k $ Which contradicts $(*)$. So $l \ge 2023$ let's suppose that $\gcd(b,c)>2023 \implies l>2023 <=> 2023<l$ From the condition again we have: $\gcd(a,b)+\gcd(a,c)+\gcd(b,c)=b+c+2023 \implies d+k+l=b+c+2023.$ Combining with $l>2023$ we have: $b+c+l>b+c+2023$ $ ...(1)$ $d+k+l>d+k+2023$ $...(2)$ From $(*)$ we have $d+k \leq b+c \implies b+c \ge d+k \implies b+c+l \ge d+k+l$, So we can subtract $(1)$ and $(2)$ $b+c+l-d-k-l>b+c+2023-d-k-2023 \implies b+c-d-k>b+c-d-k \implies b+c>b+c $ which is a contradition Since $l>2023$ and $l<2023$ failed we have that $l=2023 \implies \gcd(b,c)=2023$ It took me $3$hours to do this problem in the test (don't ask me why ) and i think i made a mistake. In the process of proving that $ l<2023$ i actually did it right but in the second part proving $l>2023$ i miscalculated something

Here is a solution I found during the contest. We will denote: $$\gcd(b,c)=d, \quad \gcd(a,b)=e, \quad \gcd(a,c)=f$$Thus there exists positive integers $x,x_1,y,z$ such that : $b=xd\quad c=x_1d \quad b=ye \quad c=zf$ With this notation the equation becomes: $$d+e+f=b+c+2023\quad (1)$$Also we know that: $d\leq b,c \quad e\leq a,b \quad f\leq a,c$ Claim 1:We cannot have $d<2023$ Proof: FTSOC suppose the opposite.Then the equation becomes: $$ \quad e+f-b-c=2023-d \quad(2)$$Which is a contradiction since in (2) we have $LHS\leq 0$ but $RHS>0$. Claim 2:We cannot have $d>2023$ Proof:Again, FTSOC suppose $d>2023$. Using earlier notation we find $x_1d=c=zf \Rightarrow d\mid \frac{x_1d}{z}=f$ and $xd=b=ey\Rightarrow d\mid \frac{xd}{y}=e$ Now notice that in (1) $d\mid b+c$ and $d\mid d+e+f$ so $d\mid 2023 \Rightarrow d\leq 2023$ .Contradiction! Therefore we must have $d=\gcd(b,c)=2023$

İ accepted a=dx, b=dy, c=dz So, gcd(a,b)=d, gcd(a,c)=d, gcd(b,c)=d 3*d=d*y+d*z+2023 3*d-d*y-d*z=2023 d*(3-y-z) = 2023 And it's given that their positive integers, so, i searched the divisors of 2023 and looked the cases. There is only 1 solution . d=2023 3-y-z=1 y+z=2 gcd(b,c)=gcd(d*y; d*z)=d=2023 I think it was simple

zaidova wrote: İ accepted a=dx, b=dy, c=dz So, gcd(a,b)=d, gcd(a,c)=d, gcd(b,c)=d 3*d=d*y+d*z+2023 3*d-d*y-d*z=2023 d*(3-y-z) = 2023 And it's given that their positive integers, so, i searched the divisors of 2023 and looked the cases. There is only 1 solution . d=2023 3-y-z=1 y+z=2 gcd(b,c)=gcd(d*y; d*z)=d=2023 I think it was simple it appeared to be a fakesolve, you can't literally assume that when $\gcd{(a,b,c)}=d$, then $\gcd{(a,b)}=\gcd{(b,c)}=\gcd{(c,a)}=d$. For a counterexample, we may take $12,15,30$. See that $\gcd{(12,15,30)}=3$, but unfortunately $\gcd{(15,30)}=15\ne 3$. so your solution is wrong.

Say that $gcd(b, c)=d$ and $b=dx$ $c=dy$ with $gcd(x, y)=1$. And say that $gcd(a, b)=\frac{dx}{p}$, $gcd(a, c)=\frac{dy}{q}$. Multiply everywhere by $pq$ to get. $d(xq+yp+pq-pqx-pqy)=2023pq$ So $xq+yp>pq(x+y-1)$. Then $WLOG: p \ge q$. So $x+y>q(x+y-1)$ $\implies$ $q=1$. And also it's easy to get $p=1$ so we are done.

JanHaj wrote: Here is a solution I found during the contest. We will denote: $$\gcd(b,c)=d, \quad \gcd(a,b)=e, \quad \gcd(a,c)=f$$Thus there exists positive integers $x,x_1,y,z$ such that : $b=xd\quad c=x_1d \quad b=ye \quad c=zf$ With this notation the equation becomes: $$d+e+f=b+c+2023\quad (1)$$Also we know that: $d\leq b,c \quad e\leq a,b \quad f\leq a,c$ Claim 1:We cannot have $d<2023$ Proof: FTSOC suppose the opposite.Then the equation becomes: $$ \quad e+f-b-c=2023-d \quad(2)$$Which is a contradiction since in (2) we have $LHS\leq 0$ but $RHS>0$. Claim 2:We cannot have $d>2023$ Proof:Again, FTSOC suppose $d>2023$. Using earlier notation we find $x_1d=c=zf \Rightarrow d\mid \frac{x_1d}{z}=f$ and $xd=b=ey\Rightarrow d\mid \frac{xd}{y}=e$ Now notice that in (1) $d\mid b+c$ and $d\mid d+e+f$ so $d\mid 2023 \Rightarrow d\leq 2023$ .Contradiction! Therefore we must have $d=\gcd(b,c)=2023$ when you say in claim 2 that $d\mid\frac{x_1d}{z}$ if this is true then $z|{x_1}$ and why is this correct?