§
3.1 Problem 8: We start with
x
2

5
y
2
= 2 and then take the remainder of both sides when divided by 5:
x
2

5
y
2
≡
x
2
≡
2
(mod 5)
.
However,
if
x
mod 5
≡
0 then
x
2
mod 5
≡
0
1 then
x
2
mod 5
≡
1
2 then
x
2
mod 5
≡
4
3 then
x
2
mod 5
≡
4
4 then
x
2
mod 5
≡
1
.
Thus it is impossible to have
x
2
≡
2
(mod 5), so
x
2

5
y
2
= 2 has no solutions.
§
3.1 Problem 12: Let’s try
a
= 2 and
b
= 4. Then
radicalbigg
a
2
+
b
2
2
=
radicalbigg
4 + 16
2
=
√
10
≈
3
.
16
>
a
+
b
2
= 3
.
Let’s try
a
= 1 and
b
= 7. Then
radicalbigg
a
2
+
b
2
2
=
radicalbigg
1 + 49
2
=
√
25 = 5
>
a
+
b
2
= 4
.
Let’s try
a
= 3 and
b
= 3. Then
radicalbigg
a
2
+
b
2
2
=
radicalbigg
9 + 9
2
=
√
9 = 3 =
a
+
b
2
= 3
.
We conjecture that the quadratic mean is always larger than or equal to the arithmetic mean.
First, if
(
a
+
b
)
/
2
<
0 then we already have
radicalbig
(
a
2
+
b
2
)
/
2
>
0
>
(
a
+
b
)
/
2. Otherwise, we work this one backwards:
radicalbigg
a
2
+
b
2
2
≥
a
+
b
2
a
2
+
b
2
2
≥
parenleftbigg
a
+
b
2
parenrightbigg
2
=
a
2
+ 2
ab
+
b
2
4
2
a
2
+ 2
b
2
≥
a
2
+ 2
ab
+
b
2
a
2

2
ab
+
b
2
≥
0
(
a

b
)
2
≥
0
.
The last line is true, with equality only when
a
=
b
. To turn this into a proof we run it the right way around:
(
a

b
)
2
≥
0
a
2

2
ab
+
b
2
≥
0
a
2
+
b
2
2
≥
a
2
+ 2
ab
+
b
2
4
=
parenleftbigg
a
+
b
2
parenrightbigg
2
2
a
2
+ 2
b
2
≥
a
2
+ 2
ab
+
b
2
radicalbigg
a
2
+
b
2
2
≥
vextendsingle
vextendsingle
vextendsingle
vextendsingle
a
+
b
2
vextendsingle
vextendsingle
vextendsingle
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 Spring '06
 CALLAHAN
 Remainder, 0 j, 1 k, 2 j, 2 3j

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