# 1991 AIME Problems/Problem 1

## Problem

Find $x^2+y^2_{}$ if $x_{}^{}$ and $y_{}^{}$ are positive integers such that $xy_{}^{}+x+y = 71$ $x^2y+xy^2 = 880^{}_{}.$

## Solution

### Solution 1

Define $a = x + y$ and $b = xy$. Then $a + b = 71$ and $ab = 880$. Solving these two equations yields a quadratic: $a^2 - 71a + 880 = 0$, which factors to $(a - 16)(a - 55) = 0$. Either $a = 16$ and $b = 55$ or $a = 55$ and $b = 16$. For the first case, it is easy to see that $(x,y)$ can be $(5,11)$ (or vice versa). In the second case, since all factors of $16$ must be $\le 16$, no two factors of $16$ can sum greater than $32$, and so there are no integral solutions for $(x,y)$. The solution is $5^2 + 11^2 = \boxed{146}$.

### Solution 2

Since $xy + x + y + 1 = 72$, this can be factored to $(x + 1)(y + 1) = 72$. As $x$ and $y$ are integers, the possible sets for $(x,y)$ (ignoring cases where $x > y$ since it is symmetrical) are $(1, 35),\ (2, 23),\ (3, 17),\ (5, 11),\ (7,8)$. The second equation factors to $(x + y)xy = 880 = 2^4 \cdot 5 \cdot 11$. The only set with a factor of $11$ is $(5,11)$, and checking shows that it is correct.

### Solution 3

Let $a=x+y$, $b=xy$ then we get the equations \begin{align*} a+b&=71\\ ab&=880 \end{align*} After finding the prime factorization of $880=2^4\cdot5\cdot11$, it's easy to obtain the solution $(a,b)=(16,55)$. Thus $$x^2+y^2=(x+y)^2-2xy=a^2-2b=16^2-2\cdot55=\boxed{146}$$ Note that if $(a,b)=(55,16)$, the answer would exceed $999$ which is invalid for an AIME answer. ~ Nafer

### Solution 4

From the first equation, we know $x+y=71-xy$. We factor the second equation as $xy(71-xy)=880$. Let $a=xy$ and rearranging we get $a^2-71a+880=(a-16)(a-55)=0$. We have two cases: (1) $x+y=16$ and $xy=55$ OR (2) $x+y=55$ and $xy=16$. We find the former is true for $(x,y) = (5,11)$. $x^2+y^2=121+25=146$.

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