1991 AIME Problem 1

Attempt Problem 1 of the 1991 AIME below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1991 AIME solutions, or check the answer key.

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1.

Find

x^2+y^2

if

x

and

y

are positive integers such that

\begin{aligned}xy+x+y&=71,\\x^2y+xy^2&=880.\end{aligned}

Answer

Answer: 146

Concepts:symmetry (algebra)system of equations quadratic

Difficulty rating: 1830

Small hint:

Let

s=x+y

and

p=xy

, and rewrite both given equations using

s

and

p

Big hint:

The two equations determine

s+p

and

sp

, so

s

and

p

are roots of one quadratic

Solution:

Let

s=x+y

and

p=xy.

The equations become

s+p=71

and

sp=880,

so

s

and

p

are the roots of

\begin{aligned}t^2-71t+880&=0,\\ (t-16)(t-55)&=0.\end{aligned}

Because

x

and

y

are positive integers,

s=16

and

p=55

; indeed,

x=5

and

y=11.

Therefore

\begin{aligned}x^2+y^2&=s^2-2p\\&=16^2-2(55)=146.\end{aligned}