Difference between revisions of "2008 AIME I Problems/Problem 4"
Mathgeek2006 (talk | contribs) m (→Solution 2) |
Mathgeek2006 (talk | contribs) m (→Solution 2) |
||
Line 14: | Line 14: | ||
We complete the square like in the first solution: <math>y^2 = (x+42)^2 + 244</math>. Since consecutive squares differ by the consecutive odd numbers, we note that <math>y</math> and <math>x+42</math> must differ by an even number. We can use casework with the even numbers, starting with <math>y-(x+42)=2</math>. | We complete the square like in the first solution: <math>y^2 = (x+42)^2 + 244</math>. Since consecutive squares differ by the consecutive odd numbers, we note that <math>y</math> and <math>x+42</math> must differ by an even number. We can use casework with the even numbers, starting with <math>y-(x+42)=2</math>. | ||
<cmath>\begin{align*}2(x+42)+1+2(x+42)+3&=244\\ | <cmath>\begin{align*}2(x+42)+1+2(x+42)+3&=244\\ | ||
− | \ | + | \Leftrightarrow x&=18\end{align*}</cmath> |
Thus, <math>y=62</math> and the answer is <math>\boxed{080}</math>. | Thus, <math>y=62</math> and the answer is <math>\boxed{080}</math>. |
Revision as of 17:27, 13 March 2015
Problem
There exist unique positive integers and
that satisfy the equation
. Find
.
Contents
Solution
Solution 1
Completing the square, . Thus
by difference of squares.
Since is even, one of the factors is even. A parity check shows that if one of them is even, then both must be even. Since
, the factors must be
and
. Since
, we have
and
; the latter equation implies that
.
Indeed, by solving, we find is the unique solution.
Solution 2
We complete the square like in the first solution: . Since consecutive squares differ by the consecutive odd numbers, we note that
and
must differ by an even number. We can use casework with the even numbers, starting with
.
Thus, and the answer is
.
Solution 3
We see that . By quadratic residues, we find that either
. Also,
, so
. Combining, we see that
.
Testing and other multiples of
, we quickly find that
is the solution.
Solution 4
We solve for x:
So is a perfect square. Since 244 is even, the difference
is even, so we try
:
,
.
Plugging into our equation, we find that , and
indeed satisfies the original equation.
Solution 5
Let for some
, substitute into the original equation to get
.
All terms except for the last one are even, hence must be even, hence let
. We obtain
. Rearrange to
.
Obviously for the right hand side is negative and the left hand side is positive. Hence
. Let
, then
.
We have . Left hand side simplifies to
. As
must be an integer,
must divide the left hand side. But
is a prime, which only leaves two options:
and
.
Option gives us a negative
. Option
gives us
, and
, hence
.
See also
2008 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.