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2003 Poland - Second Round
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4
Part of
2003 Poland - Second Round
Problems
(1)
Sum of two squares, equation
Source: 54 Polish MO 2003 Second Round - Second Day Problem 4
4/21/2018
Prove that for any prime number
p
>
3
p > 3
p
>
3
exist integers
x
,
y
,
k
x, y, k
x
,
y
,
k
that meet conditions:
0
<
2
k
<
p
0 < 2k < p
0
<
2
k
<
p
and
k
p
+
3
=
x
2
+
y
2
kp + 3 = x^2 + y^2
k
p
+
3
=
x
2
+
y
2
.
number theory
combinatorics
Combinatorial Number Theory
Poland
pigeonhole principle
prime numbers