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2009 JBMO Shortlist
4
p_1^2+ p_2^2+ ... + p_{12}^2 = p_{13}^2
p_1^2+ p_2^2+ ... + p_{12}^2 = p_{13}^2
Source: JBMO 2009 Shortlist N4
October 14, 2017
JBMO
number theory
Sum of powers
Problem Statement
Determine all prime numbers
p
1
,
p
2
,
.
.
.
,
p
12
,
p
13
,
p
1
≤
p
2
≤
.
.
.
≤
p
12
≤
p
13
p_1, p_2,..., p_{12}, p_{13}, p_1 \le p_2 \le ... \le p_{12} \le p_{13}
p
1
,
p
2
,
...
,
p
12
,
p
13
,
p
1
≤
p
2
≤
...
≤
p
12
≤
p
13
, such that
p
1
2
+
p
2
2
+
.
.
.
+
p
12
2
=
p
13
2
p_1^2+ p_2^2+ ... + p_{12}^2 = p_{13}^2
p
1
2
+
p
2
2
+
...
+
p
12
2
=
p
13
2
and one of them is equal to
2
p
1
+
p
9
2p_1 + p_9
2
p
1
+
p
9
.
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