A number theory problem from the British Math Olympiad
by Rainbow1971, Mar 28, 2025, 8:39 PM
I am a little surprised to find that I am (so far) unable to solve this little problem:
I set
. If 
is an integer, then 
is at least rational, so that 
must be a perfect square then. Using Conway's topograph method, I have found out that the smallest non-negative pairs 
for which this happens are 
and 
, and that, for every such pair 
, the "next" such pair can be calculated as
The eigenvalues of that matrix are irrational, however, so that any calculation which uses powers of that matrix is a little cumbersome. There must be an easier way, but I cannot find it. Can you?
Thank you.
Quote:
Let 
be an integer. Show that, if is an integer, then it is a perfect square.
be an integer. Show that, if 
is an integer, then it is a perfect square.I set
. If is an integer, then 
is at least rational, so that 
must be a perfect square then. Using Conway's topograph method, I have found out that the smallest non-negative pairs 
for which this happens are 
and 
, and that, for every such pair 
, the "next" such pair can be calculated as
The eigenvalues of that matrix are irrational, however, so that any calculation which uses powers of that matrix is a little cumbersome. There must be an easier way, but I cannot find it. Can you?Thank you.
be a triangle, and let 
and 
be points on segment 
symmetric with respect to the midpoint of 
denote the circle passing through 
at 
.
denote the circle passing through 
at 
.
)
.
,![\[
|A \cup B| = |A| + |B| - |A \cap B|.
\]](http://latex.artofproblemsolving.com/6/f/3/6f3e15b9f2bdb4331acc41fa09951927b82d5014.png)
are counted twice—once in 
and once in 
—![\[
|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|.
\]](http://latex.artofproblemsolving.com/6/c/9/6c903bc32ca8e0072bc277d0f441b7ab7bfa3eca.png)
,![\[
\left| \bigcup_{i=1}^n A_i \right| = \sum_{k=1}^n (-1)^{k+1} \sum_{1 \leq i_1 < \dots < i_k \leq n} |A_{i_1} \cap \dots \cap A_{i_k}|.
\]](http://latex.artofproblemsolving.com/d/f/b/dfbfeb1be8ae686a72f94d98c0a9545baf7c6094.png)
that belongs to at least one of the sets. We need to ensure each element is counted exactly once.![\[
m(x) = \text{Number of sets } A_i \text{ that contain } x.
\]](http://latex.artofproblemsolving.com/7/5/2/752600a41711228882174ede71b5c691681dfeb1.png)
and 
are divisible by 
,
,
.
![\[
|A| = \left\lfloor \frac{100}{2} \right\rfloor = 50, \quad |B| = \left\lfloor \frac{100}{3} \right\rfloor = 33, \quad |C| = \left\lfloor \frac{100}{5} \right\rfloor = 20
\]](http://latex.artofproblemsolving.com/e/d/c/edcadd78cbc24e4a2cf63e321f0b3882a570c65e.png)
![\[
|A \cap B| = \left\lfloor \frac{100}{6} \right\rfloor = 16, \quad |A \cap C| = \left\lfloor \frac{100}{10} \right\rfloor = 10, \quad |B \cap C| = \left\lfloor \frac{100}{15} \right\rfloor = 6
\]](http://latex.artofproblemsolving.com/7/4/5/745c9fa1f79f7c5225b905fc989c0e1f758e7f93.png)
![\[
|A \cap B \cap C| = \left\lfloor \frac{100}{30} \right\rfloor = 3
\]](http://latex.artofproblemsolving.com/e/e/b/eebf2c896136ef1c5193cdf26481e2fc2fbb0816.png)
![\[
|A \cup B \cup C| = 50 + 33 + 20 - 16 - 10 - 6 + 3 = 74
\]](http://latex.artofproblemsolving.com/5/1/1/51166af1a180dadbb4afedc762bcd952ae33b57b.png)
numbers between 
and measurable sets 
,![\[
\mu \left( \bigcup_{i=1}^n A_i \right) = \sum_{k=1}^n (-1)^{k+1} \sum_{1 \leq i_1 < \dots < i_k \leq n} \mu(A_{i_1} \cap \dots \cap A_{i_k}).
\]](http://latex.artofproblemsolving.com/8/6/9/8693c12d0a177189733a29628374e7a4c51b5b7c.png)
represent the number of derangements of ![\[
D_n = n! \sum_{k=0}^n \frac{(-1)^k}{k!},
\]](http://latex.artofproblemsolving.com/b/2/f/b2f5e9eda7c3683161636aec79924d453e445c9e.png)
and circumcenter 
.
and 
be points on the line segments 
is a parallelogram. Prove that 
.
and 
satisfies the following recursion formulas: 
,
.
,
and 
.
is prime, 
with 
and 
interesting if 
of 
and 
is also a divisor of 
that satisfy 
for all 
March 2025