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PEN H Problems
78
H 78
H 78
Source:
May 25, 2007
Diophantine Equations
pen
Problem Statement
Let
x
,
y
x, y
x
,
y
, and
z
z
z
be integers with
z
>
1
z>1
z
>
1
. Show that
(
x
+
1
)
2
+
(
x
+
2
)
2
+
⋯
+
(
x
+
99
)
2
≠
y
z
.
(x+1)^{2}+(x+2)^{2}+\cdots+(x+99)^{2}\neq y^{z}.
(
x
+
1
)
2
+
(
x
+
2
)
2
+
⋯
+
(
x
+
99
)
2
=
y
z
.
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