WakeUp wrote: Let $a,b,c$ be positive integers. Prove that it is impossible to have all of the three numbers $a^2+b+c,b^2+c+a,c^2+a+b$ to be perfect squares. $a^2+b+c$ perfect square $\implies$ $b+c\ge 2a+1$ $b^2+c+a$ perfect square $\implies$ $c+a\ge 2b+1$ $c^2+a+b$ perfect square $\implies$ $a+b\ge 2c+1$ And adding these three inequalities, we get $0\ge 3$, impossible. Hence the result.

another way: let $a\ge b\ge c$. then $2a\ge b+c\ge 2a+1$ which is impossible.

Let $a\ge b\ge c$ Then $a^2<a^2+b+c\le a^2+2a<(a+1)^2$

Suppose that $a^2+b+c, b^2+c+a, c^2+a+b$ are all perfect squares, let's say \begin{align} a^2+b+c&=x^2\\ b^2+a+c&=y^2\\ c^2+a+b&=z^2 \end{align} Then the discriminants of these three quadratic equations in $x,y$ and $z$, respectively: $\Delta_1:=0^2-4(a^2+b+c),\Delta_2:=0^2-4(b^2+a+c),\Delta_3:=0^2-4(c^2+a+b)$ are all nonnegative, or $\Delta_i\ge 0,\ i=1,2,3$. Therefore $\begin{align*} \Delta_1+\Delta_2+\Delta_3&=-4(a^2+b^2+c^2+2a+2b+2c)\\ &\ge 0\\ &\Longrightarrow a^2+b^2+c^2+2a+2b+2c\le 0\\ &\Longrightarrow (a+1)^2+(b+1)^2+(c+1)^2\le 3$ Contradiction, since $a,b,c$ are all positive integers.

Is this trivial problem from Apmo 2011? Are you missing something? Also, is the ban time over?

yes, the ban time is over. and also this problem is trivial as you say, the official solution is like the one pco wrote.

A very disappointing problem. That was the shortest solution I've ever written on any olympiad (only 3 lines!) and also the quickest.

funny thing is, I couldn't come up with this solution for 4 hours when i took it officially. Fail point for me.

Hi PCO, Would you please let me know as to how you deduced $a^2+b+c$ a perfect square $\implies$ $b+c\ge 2a+1$

Note that $a^2+b+c$ is necessarily greater than $a^2$, so if it is a perfect square, it must be at least $(a+1)^2$. Thus \[a^2+b+c\geq a^2+2a+1\implies b+c\geq 2a+1.\]

thanks djmathman

Another idea: Assume that all 3 are perfect squares Let $a^2+b+c = x^2, a+b^2+c = y^2, a+b+c^2 = z^2$ where x,y,z must be integers. Then $x^2 - y^2 = a^2-b^2-a+b = (a+b)(a-b)-(a-b) = (a+b-1)(a-b)=(x+y)(x-y)$. It is easy to see that $x = \frac{2a-1}{2}, y = \frac{2b-1}{2}$. Thus we see that $a = \frac{2x+1}{2}, b = \frac{2y+1}{2}$ and by symmetry $c = \frac{2z+1}{2}$. Replacing this into $a^2+b+c$ we see that $(\frac{2x+1}{2})^2 + \frac{2y+1}{2} + \frac{2z+1}{2} = \frac{4x^2+4x+4y+4z+1+2+2}{4} = x^2+x+y+z+1+\frac{1}{4}$ which cannot be a perfect square as perfect squares are integers, which is a contradiction.

hypogean

https://artofproblemsolving.com/community/c3046h1349244_the_system_of_diophantus

This has got to be the easiest APMO problem ever: Say $a \geq b \geq c \implies a^2+b+c \leq a^2+2a \implies a^2<a^2+b+c <(a+1)^2$.

Trivial but it took me more than 30 minutes to solve. WakeUp wrote: Let $a,b,c$ be positive integers. Prove that it is impossible to have all of the three numbers $a^2+b+c,b^2+c+a,c^2+a+b$ to be perfect squares. Firstly if $a=b$ or $b=c$ or $c=a$ then either $a^2+b+c$ or $b^2+c+a$ or $c^2+a+b$ is of the form $k^2-1$ hence, not a perfect square. So $a\neq b\neq c$ and WLOG let $a>b>c$, hence we get that $a^2<a^2+b+c<a^2+2a<a^2+2a+1=(a+1)^2$ and as $a^2+b+c$ lies strictly between two perfect squares, hence cannot be a perfect square. Hence, $a^2+b+c,b^2+c+a,c^2+a+b$ cannot be perfect squares simultaneously. $\blacksquare$

ubermensch wrote: This has got to be the easiest APMO problem ever: You'll soon find out the most trivial problem

I solved this problem in 10 seconds approximately, this is more trivial than school problems By this method ubermensch wrote: This has got to be the easiest APMO problem ever: Say $a \geq b \geq c \implies a^2+b+c \leq a^2+2a \implies a^2<a^2+b+c <(a+1)^2$.

Quite easy. Since $a,b,c$ are positive, for $a^2+b+c$ to be a perfect square it must be at least $(a+1)^2=a^2+2a+1$, so we have $b+c \geq 2a+1$. Similar inequalities hold for the other expressions. Summing up, we get $2a+2b+2c \geq 2a+2b+2c+3$, which is clearly false. Thus the desired is impossible.

If $a^2+b+c$ is a perfect square, then $a^2+b+c>a^2$, so $a^2+b+c\ge(a+1)^2$ implying that $b+c\ge2a+1$. Adding up the cyclic variations of this, we obtain that $0\ge3$, so no solutions. $\square$

$a^2+b+c \implies b+c \ge (2a+1)$, which $\sum_{cyc} a^2+b+c \implies 2a+2b+2c \ge 2a+2b+2c+3 \implies 0 \ge 3$, which is obviously not true.

ubermensch wrote: This has got to be the easiest APMO problem ever: Say $a \geq b \geq c \implies a^2+b+c \leq a^2+2a \implies a^2<a^2+b+c <(a+1)^2$. how on earth could I have solved an APMO question in 15 minutes ??!!

Suppose they are. Then $b+c\ge 2a+1, a+c\ge 2b+1$, $a+b\ge 2c+1$. Adding gives a contradiction.

For positive integers $m,n,$ we know $n^2-m^2\ge 2m+1$ if $n>m.$ Hence, if $a^2+b+c,b^2+c+a,c^2+a+b$ are perfect square, $b+c\ge 2a+1,c+a\ge 2b+1,$ and $a+b\ge 2c+1.$ Adding yields $0\ge 3,$ a contradiction. $\square$

Solved with cxyerl and cj13609517288 (VA gang fr) Assume that they are perfect squares. WLOG $a \geq b \geq c$. We have $a^2+b+c \geq (a+1)^2$ which means $b+c \geq 2a+1$. However, we have $2a \geq b+c$, which is clearly a contradiction. $\blacksquare$

AFTSOC they all are perfect square at the same time. So we get \begin{align*} (a+1)^2 \leq a^2+b+c \\ (b+1)^2 \leq b^2+a+c \\ (c+1)^2 \leq c^2+a+b \end{align*}Rearranging gives \[ b+c\geq 2a+1, \qquad a+c\geq 2b+1, \qquad a+b\geq 2c+1 \]Adding these three equation gives $0\geq 3$, contradiction!

Bruh, I can't believe this took me more than 40 minutes to come up with, maybe I should lower my expectations for P1s. Sps there are positive integers $x_1,x_2,x_3$ such that $_{cyc}(a^2+b+c=x_1^2)$. It's clear that: $_{cyc}(x_1 \ge a+1)$. But summing up the equations, we have: $\sum_{cyc}((a+1)^2-x_1^2)=3$ , but then $0 \ge 3$, a contradiction.

I use a bounding proof. Assume WLOG that a, b, c are in a not strictly decreasing order. Then consider the largest value which is a^2+b+c. This is greater than or equal to a^2, however, it is less than or equal to a^2 + 2a, which is less than (a+1)^2. So the square has to be a^2, then b+c =0 which is not true at all for positive integers. So the conclusion is that no solutions exist

WakeUp wrote: Let $a,b,c$ be positive integers. Prove that it is impossible to have all of the three numbers $a^2+b+c,b^2+c+a,c^2+a+b$ to be perfect squares.

By the way, I wondering if there is a modulo proof for this problem?

Rizztastic problem with bounding! Clearly b+c >= 2a+1 for it to be square, and summing this fanumtastic inequality gives a contradiction! Therefore not all of the numbers can be squares!

apmoball!!

Without loss of generality assume \(\max\{a, b, c\} = a\), so \(b, c \leq a\). We claim that \[a^2 < a^2 + b + c < (a + 1)^2 = a^2 + 2a + 1\]and thus \(a^2 + b + c\) cannot be a perfect square. The left inequality is trivial. For the right inequality, note that \[a^2 + b + c \leq a^2 + a + a < a^2 + 2a + 1\]as desired. The problem is solved.

WLOG $a \ge b \ge c$, then $a^2 < a^2 + b + c \le a^2 + a + a = a^2 + 2a < a^2 + 2a + 1 = (a+1)^2$, hence $a^2 + b + c$ cannot be a perfect square.