Introductory Number Theory. Homework #1
Some solutions and grading notes.
1. Textbook, #1.1 (p. 11)
One can make a list of triangular numbers and squares, and see what numbers appear in both, but that might
take a lot of time and actually tell us little. The question is really: For what natural numbers
n, m
do we
have
n
(
n
+ 1)
2
=
m
2
.
(1)
Another way of writing this is as
n
(
n
+ 1) = 2
m
2
. The factor of 2 suggests dividing the quest into two cases,
n
odd,
n
even. Since the textbook left it as a somewhat open ended question, I’ll just give you my solution
and leave it up to you, and your discretion, to verify that I covered all possibilities, and that all works.
I am
not assuming that any of you got as complete a solution
; though it will be nice if some of you did.
It took me a while to figure all this out; things like these require work. They don’t drop like manna from
heaven.
Case
n
odd.
In my solution (your’s could be simpler!) I generate inductively two sequences of numbers,
{
‘
ν
}
and
{
r
ν
}
as follows.
‘
1
= 1. Assuming
‘
ν
defined for some
ν
≥
1 I define
r
ν
=
‘
ν
+
p
2
‘
2
ν

1
,
and then
‘
ν
+1
=
r
ν
+
p
2
r
2
ν
+ 1
.
It is not at all obvious at first sight that all these numbers are integers, but an inductive proof shows they
are. All solutions of the equation (1) for
n
odd are given by
n
= 2
‘
2
ν

1
, ν
= 1
,
2
,
3
, . . .
;
an infinity of solutions. The corresponding
m
is of course found by taking the square root of
n
(
n
+1)
/
2. Here
are the few first ones:
•
with
ν
= 1, have
‘
1
= 1, so
n
= 1. We get the triangular number
1
, which is also a square number.
•
I’ll omit the subscript from
‘, r
.
With
‘
= 1 we get
r
= 1 +
√
2
·
1
2

1 = 2; thus the next
‘
is
‘
= 2 +
√
2
·
2
2
+ 1 = 5, thus
n
= 2
·
5
2

1 = 49, we get the triangular/square number
49
·
50
2
= 1225 = 35
2
.
•
The next value of
r
is
r
= 5 +
√
49 = 12; then
‘
= 12 +
√
289 = 29; this gives
n
= 1681, and we get the
triangular/square number
1681
·
1682
2
= 1
,
413
,
721 = 1189
2
.
•
The next
r
is given by
r
= 29+
√
1681 = 70, then
‘
= 70+
√
9801 = 169. This gives a value of
n
= 57
,
121
and the triangular/square number
57121
·
57121
2
= 1
,
631
,
432
,
881 = 40
,
391
2
.
And so it goes. There is an infinity of triangular/square numbers with odd
n
.
The case
n
even.
Here we simply need to swap the
‘
’s and
r
’s from the odd case. With the same definition,
all the triangular/square numbers with
n
even are obtained by taking
n
= 2
r
2
ν
,
ν
= 1
,
2
, . . .
. Here are the
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first few ones, recalling that the first few values of
r
were 2
,
12
,
70:
n
= 2
·
2
2
= 8
gives the triang/sq. number
8
·
9
2
= 36 = 6
2
,
n
= 2
·
12
2
= 288
gives the triang/sq. number
288
·
289
2
= 41
,
616 = 204
2
,
n
= 2
·
70
2
= 9800
gives the triang/sq. number
9800
·
9801
2
= 48
,
024
,
900 = 6930
2
.
There is also an infinity of such numbers with even
n
.
Grading:
10 points for finding the square/triangular number after 36
AND
explaining how you found it
*
,
showing some effort in trying to find some more, and reaching no unwarranted conclusions.
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 Summer '11
 STAFF
 Natural number, Prime number, ν, Pythagorean triple, JoAnne Growney

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