Difference between revisions of "2008 AIME I Problems/Problem 4"
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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>. | ||
− | < | + | <center><math>\begin{align*}2(x+42)+1+2(x+42)+3&=244\\ |
− | \Rightleftarrow x&=18\end{align*}</ | + | \Rightleftarrow x&=18\end{align*}</math></center> |
− | Thus, <math>y=62</math>, | + | Thus, <math>y=62</math>, which works, so <math>x+y=80</math>. |
===Solution 3=== | ===Solution 3=== |
Revision as of 22:45, 24 March 2008
Problem
There exist unique positive integers and
that satisfy the equation
. Find
.
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. Sine
, 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, , which works, so
.
Solution 3
, we see that
; by quadratic residues, we find that either
. Also,
,
, and so
. Combining, we see that
.
Testing and other multiples of
, we quickly find that
is the solution.
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 |