Number Theory
Review for Final Exam
The exam is scheduled for Monday, May 1 from 8-10 in EN 2100.
Powers modulo p
Be able to :
• Use Fermat’s Little Theorem
• Compute the order of integers mod p
• Prove and use the order theorem
• Compute order of xk mod p knowing the order of x.
Let p be a prime number, and let a be an integer. Fermat’s Little theorem asserts that
ap ≡ a mod p.
(1)
However, we can make this a bit more specific. If p|a, then for each positive integer k we have ak ≡ 0 mod p. If
p doesn’t divide a, then there exists an integer x with ax ≡ 1 mod p (here’s why: Since p doesn’t divide a, a
and p are relatively prime. So there exist integers x and y with ax + py = 1. Reading mod p we get
ax ≡ 1 mod p. ), and then by multiplying (1) by x we have
ap−1 ≡ ap−1 ax ≡ ax ≡ 1 mod p.
Thus, when p 6 |a, we have ap−1 ≡ 1 mod p.
For a 6≡ 0 mod p, we defined the order of a mod p to be the smallest positive integer k such that ak ≡ 1 mod p.
We denote this by ordp (a). For example, ord7 (1) = 1, ord7 (6) = 2, ord7 (3) = 6. If ordp (x) = 2 then x2 ≡ 1 mod p,
and x 6≡ 1 mod p. So, x ≡ −1 mod p.
The Order theorem asserts that if p is a prime and a 6≡ 0 mod p, then an ≡ 1 mod p if and only if ordp (a)
divides n.
The proof of the order theorem is in two parts. First suppose that ordp (a) divides n. Then n = ordp (a)k for
some integer k, and an ≡ aordp (a)k ≡ (aordp (a) )k ≡ (1)k ≡ 1 mod p, as desired. Second suppose that n ≡ 1 mod p.
Let d = ordp (a). We wish to show that d|n. To do this we use long-division to write n = dq + r where 0 ≤ r < d.
Now, 1 ≡ an ≡ adq+r ≡ adq ar ≡ (ad )q ar ≡ (1)r ar ≡ ar mod p. Then ar ≡ 1 mod p. However, 0 ≤ r < d, and d is
the smallest positive integer such that ad ≡ 1 mod p. Thus, we must have r = 0. This shows that d|n, as desired.
We have the following consequences of the order theorem:
• By Fermat’s Little Theorem, ap−1 ≡ 1 mod p. Thus, the order theorem implies that ordp (a) divides p − 1.
• If a has order d mod p, then ak has order d/gcd(d, k). (This can be argued as follows: First note that
(ak )d/gcd(d,k) = (ad )k/gcd(d,k) ≡ 1k/gcd(d,k) . Thus, ordp (ak ) divides d/gcd(d, k). Next suppose that
(ak )` ≡ 1. Then ak` ≡ 1. By the order theorem, d|k`. Thus, d/gcd(d, k)|k/gcd(d, k)`. Since
d/mboxgcd(d, k) and kgcd(d, k) are relatively prime, d/gcd(d, k) divides `. Therefore ordp (ak ) ≥
d/gcd(d, k). This proves the result.
• ak ≡ a` mod p if and only if k ≡ ` mod ordp (a).
An integer x is a primitive root mod p provided p 6 |x, and ordp (x) = p − 1. For example, the primitive roots (in
{1, 2, 3, 4}) mod 5 are {2, 3}. We assumed the known result that for every prime p, there exists a primitive root, w,
mod p. Note that since w is primitive no two of {1, w1 , w2 , . . . , wp−2 } are congruent modulo p (for if wi ≡ wj mod p,
then wj−i ≡ modp, and then the order of w is not p − 1.) Thus, the set {1, w1 , w2 , . . . , wp−2 } when reduced mod
p is the same as the set {1, 2, 3, . . . , p − 2}. For example, with p = 7, and w = 3 we get {1, 3, 32 ≡ 2 mod 7, 33 ≡
6 mod 7, 34 ≡ 4 mod 7, 35 ≡ 5 mod 7, 36 ≡ 1 mod 7}. Thus, ord7 (3) = 6, ord7 (2) = ord7 (32 ) = 6/gcd(6, 2) = 3,
ord7 (6) = ord7 (33 ) = 6/gcd(6, 3) = 2, ord7 (4) = ord7 (34 ) = 6/gcd(6, 4) = 3, ord7 (5) = ord7 (35 ) = 6/gcd(6, 5) = 6.
In particular, we see that the number of primitive roots mod p {1, w, w2 , . . . , wp−2 } are precisely those of the
form wj where j and p − 2 are relatively prime. For example, for p = 7, and w = 3., the primitive roots mod p are
w1 and w5 (i.e. 3 and 5). Remember, we let φ(n) denote the number of integers in 1, 2, . . . , n − 1 that are relatively
prime to n. Thus, the number of primitive roots mod p is φ(p − 2).
Problems:
1.1 I have a prime p and an integer a such that a9 ≡ 1 mod p and a12 ≡ 1 mod p. What can you say about
the order of a? Use the Order theorem to explain your answer. Let d be the order of a. By the order
theorem d|9 and d|12. Thus, d|gcd(9, 12) = 3. Hence d = 1 or d = 3.
1.2 I’ve checked that the order a mod 13 is at least 5. What are the possibilities for the order of a. Why? By a
consequence of the order theorem, the order of a divides 13 − 1 = 12. Since the order is at least
5, then the order is either 6 or 12.
1.3 True or false: there is an integer k whose order mod 19 is 7?
(Hint: you needn’t do lots of computations) In mod 19 the order always divides 19 − 1 = 18. Since 7
doesn’t divide 18. The answer is FALSE.
1.4 True or false: there is an integer k whose order mod 19 is 6? True. We know that there is a primitive
19th root, w. So w has order 18. Now w3 has order 18/gcd(18, 3) = 6.
1.5 (a) Find (by trial and error) a primitive root mod 11. 2 works.
(b) Find all primitive roots mod 11 in {1, 2, . . . , 10}. We look at the 2i mod 11 for which 1 ≤ i ≤ 10 and
gcd(i, 10) = 1. Thus the primitive roots mod 11 are: 2, 23 = 8, 27 ≡ 7, 29 ≡ 6
(c) Find the ord11 (x) for each x in {1, 2, . . . , 10}.
Order of 1 ≡ 210 is 10/gcd(10, 0) = 1
Order of 2 = 21 is 10/gcd(10, 1) = 10
Order of 4 = 22 is 10/gcd(10, 2) = 5
Order of 8 = 23 is 10/gcd(10, 3) = 10
Order of 5 ≡ 24 is 10/gcd(10, 4) = 5
Order of 10 ≡ 25 is 10/gcd(10, 5) = 2
Order of 9 ≡ 26 is 10/gcd(10, 6) = 5
Order of 7 ≡ 27 is 10/gcd(10, 7) = 10
Order of 3 ≡ 28 is 10/gcd(10, 8) = 5
Order of 6 ≡ 29 is 10/gcd(10, 9) = 10
1.6 How many primitive roots mod 13 are there? φ(13 − 1) = φ(12) = 4 because there are 4 numbers between
1 and 12 relatively prime to 12 (namely: 1,5,7,11)
Quadratic Residues
Throughout this section p is a prime, and a and b are integers. Be able to:
• Use properties of QR and NR’s
• Determine QR’s and NR’s mod small primes
The integer a is a quadratic residue mod p provided a 6≡ 0 mod p, and there is an integer x with x2 ≡ a mod p.
For example, 2 is a quadratic residue mod 7 because 32 ≡ 2 mod 7. We say a is QR mod p if a is a quadratic residue
mod p.
The integer a is a quadratic non-residue mod p provided a 6≡ 0 mod p, and there is not an integer x with
x2 ≡ a mod p. We write a is NR mod p to indicate that a is a quadratic non-residue mod p.
If p|a, then a is neither QR or NR.
Let w be a primitive root mod p. As we saw in the last section, each of the numbers 1,2,. . . , p−1 is congruent mod
p to exactly one of 1, w, w2 , . . . , wp−2 . Clearly, each of 1, w2 , w4 , . . . , wp−3 is a QR (because (wi )2 ≡ w2i mod p).
Can a w2j+1 be a QR? No! Here’s why. Suppose that w2j+1 ≡ x2 mod p for some x. Then x ≡ wi for some i, and we
have w2j+1 ≡ w2i mod p. This requires that 2j + 1 ≡ 2i mod p − 1 (by a consequence of the order theorem).
In particular this says that 2j + 1 − 2i is even (because p − 1 is even and divides 2j + 1 − 2i). But this is certainly
not the case–we’ve reached a contradiction. Thus w, w3 , . . . , wp−2 are all NR’s mod p. Therefore we have proven
If p is an odd prime, then the number of QR’s mod p (in {1, 2, . . . , p − 1}) is (p − 1)/2, and the number
of NR’s is also (p − 1)/2.
In addition, note that wi is a QR mod p if and only if i is even, and wi is a NR mod p if and only if i is odd.
Since even + even = even, even + odd = odd, and odd + odd = even, we have the following rules: QR · QR = QR,
QR · N R = NR, N R · QR = NR, and N R · N R = QR.
It is useful to know when −1 is a QR mod p. Certainly, if p = 2, then −1 ≡ 1 (mod 2), and −1 is a QR mod
2. Now assume that p is odd. Suppose −1 is a QR mod p. Then there exists an integer x with x2 ≡ −1 mod p.
Squaring gives that x4 ≡ 1 mod p. Thus, the order of x mod p divides 4. If the order of x mod p is 1 or 2, then
x2 ≡ 1 mod p, rather than x2 ≡ −1 mod p. Thus the order of x mod p is 4, and 4|(p − 1) (because Fermat’s Little
theorem and the order theorem). This means that p ≡ 1 mod 4. Now suppose that p ≡ 1 mod 4. Let w be a
primitive root mod p. Let k be the integer (p − 1)/4, and let x = wk . Then x4 ≡ wp−1 ≡ 1 mod p (by Fermat’s
Little theorem). This means that (x2 )2 ≡ modp, and thus either x2 ≡ 1 mod p or x2 ≡ −1 mod p. The former
would say that w has order less than (p − 1)/2 which is impossible. Thus, x2 ≡ −1 mod p, and −1 is a QR mod p.
We have shown:
−1 is a QR mod p if and only if p = 2 or p ≡ 1 mod 4.
Problems
2.1 Determine all the QR’s mod 11, and all the NR’s mod 11. Solution 1: We look at 12 , 22 , 32 , 42 , 52 and
reduce each mod 11 to get: 1,4,9,5,3.
Solution 2: By Problem 1.5, 2 is a primitive root mod 11. So we look at 20 , 22 , 24 , 26 , 28 and
reduce mod 11 to get: 1,4,5,9, 3
2.2 How many QR’s (in {1, 2, . . . , 96}) mod 97 are there? Answer: (97 − 1)/2 = 48.
2.3 It is known that a is a NR mod 97.
(a) Is a3 a NR mod 97?
N R · N R · N R = QR · N R = N R. Yes.
(b) Is −a a NR mod 97? 97 ≡ 1 mod 4. So −1 is a QR. a = −1 · a = QR · N R = N R. So answer is yes!
2.4 Let p be an odd prime. Show that there exists an integer x such that x3 ≡ −1 mod p and x2 6≡ 1 mod p if and
only if p ≡ 1 mod 6. First suppose that p ≡ 1 mod 6. Let w be a primitive root mod p. Consider
x = w(p−1)/6 . Then (x3 )2 ≡ wp−1 ≡ 1 mod p. Thus x3 ≡ 1 mod 6 or x3 ≡ −1 mod p. The former can’t
be, lest w’s order would be less than 3(p − 1)/6 < p. Also x2 = (w(p−1)/6 )2 = w(p−1)/3 which can’t be
1 mod p, since the order of w is p − 1.
Conversely, suppose that there exists an integer x such that x3 ≡ −1 mod p and x2 6≡ 1 mod p.
Squaring the first equation shows that x6 ≡ 1 mod p. So by the order theorem the order of x
divides 6. Since x2 and x3 aren’t 1 mod p, the order of x must be 6. By the order theorem 6
divides p − 1. Thus, p ≡ 1 mod 6.
Legendre Symbol
Throughout p is an odd prime and a and b are integers. Be able to:
• Use and prove Euler’s theorem
• Use Gauss’s Lemma to compute Legendre symbols
• Use the Law of Quadratic Reciprocity
We defined a useful symbol, called as the Legendre symbol.

 0 if p divides a
a
1 a is a QR mod p
:=

p
−1 a is a NR mod p.
Euler’s theorem asserts:
If p is an odd prime and a is an integer, than a(p−1)/2 ≡
a
p
This is proven as follows: if p|a, then a(p−1)/2 ≡ 0 mod p, and
.
a
p
= 0. Otherwise, let x = a(p−1)/2 . Then
x2 ≡ ap−1 ≡ 1 mod p (by Fermat’s Little theorem). Thus x ≡ ±1 mod p. Let w be a primitive root mod p. We
showed in the previous section that x is QR if x ≡ w2i mod p for some i, and x is NR if x ≡ w2j+1 mod p for some j.
Case 1: x is QR. So x ≡ w2i mod p. So
x(p−1)/2 ≡ (wp−1 )i ≡ 1i ≡ 1 mod p
(by Fermat’s Little theorem). Thus, x(p−1)/2 ≡ 1 mod p, and ap = 1 .
Case 2: x is NR. So x ≡ w2j+1 mod p. Thus, x(p−1)/2 ≡ w(p−1)(2j+1)/2 mod p. Since 2j +1 is odd, (p−1)
does not
(p−1)/2
(p−1)/2
divide (p−1)(2j+1)/2, and thus (since ordp (w) = p−1), x
6≡ 1 mod 2. Therefore, x
≡ −1 ≡ ap mod p,
as desired.
This gives us the following results for the Legendre symbol:
• ap ≡ a(p−1)/2 mod p.
• If a ≡ b mod p, then
•
a
p
b
p
=
ab
p
a
p
=
b
p
.
Not to be outdone, Gauss entered the picture, and gave another way of computing
a
p
, when p 6 |a. He looked
p−1
2 a
at the remainders when the numbers a, 2a, . . .
where divided by p. He let s be the number of these remainders
that were greater than p/2. Then Gauss’ Lemma asserts that
a
= (−1)s .
p
As an application of this, let a = 2. Then the remainders when 2, 4, 6, . . . (p − 1)/2 are divided by p are
2, 4, 6, 8, . . . , (p − 1)/2. Of these remainders, bp/4c less than p/2. Thus, s = (p − 1)/2 − bp/4c. With a bit of
arithmetic we can show that
1 if p ≡ 3 mod 8 or p ≡ 5 mod 8
(−1)s =
−1 if p ≡ 1 mod 8 or p ≡ 7 mod 8.
Problems
5
3.1 Use Gauss’ Lemma to find 11
We look at the remainder of 5, 2 · 5, 3 · 5, 4 · 5 and
5 · 5 2when divided
5
by 11. This are: 5, 10, 4, 9, 3. Two of these are larger than 11/2. Thus 11 = (−1) = 1.
3.2 For which of the following primes is 2 a QR?
2, 3, 7, 11, 13, 17, 19 2, 7, 17,23
3
a
3.3 It is known that 97
= −1. Find −4a
. Note 97 ≡ 1 mod 4. So
97
1 · (−1)3 = −1
−1
97
= 1. Thus
−4a3
97
=
−1
97
[
a
97
3
] =
Law of Quadratic Reciprocity and Sums of Squares
Be able to
• Use Law of Quadratic Recripocity to compute Legendre symbols and determine whether or not a quadratic
equation (mod p) is solvable
• Determine whether or not a given integer is a sum of squares
• Write a given prime p (which is ≡ 1 mod 4) as a sum of squares using our reduction algorithm
Using the fact that the product of Legendre symbols (mod p) is the Legendre symbol
of the product, we can
reduce the problem of determining any Legendre symbol into the problem of determing pq in the case that p and
q are distinct primes.
Using Gauss’ Lemma one can prove the following amazing result:
p
Law of Quadratic Reciprocity Let p and q be distinct odd primes. Then pq
is 1 if either p or
q
q is congruent to 1 mod 4, and is −1 if both p and q are congruent to 3 mod 4.
Our final topic for the semester concerned sums of squares. Note that if the odd prime p is a sum of squares,
then there exist integers x and y such that x2 + y 2 = p ≡ 0 mod p, or equivalently x2 ≡ −y 2 mod p. Thus either
x ≡ y ≡ 0 mod p or ,
2 2 2 −y
−1
y
−1
x
=
=
=
.
1=
p
p
p
p
p
If the former, then p2 |(x2 + y 2 ), which is impossible. So the latter holds, and thus −1 is a QR mod p, and as we’ve
seen this implies that p ≡ 1 mod 4. The converse is also true, namely if p ≡ 1 mod 4, then p is a sum of squares.
Our argument for this went as follows. Assume that p ≡ 1 mod 4. Then −1 is a QR mod p. This means that there
exist integers x and k such that x2 + 1 = kp. Note that this can be viewed as x2 + 12 = kp. Thus, some multiple of
p is a sum of squares. We now show how we can make that multiple of p be just p.
Suppose that we have positive integers u and v with u2 + v 2 = sp for some integer s > 1. We’ll show how we can
find positive integers u0 , v 0 and s0 with u02 + v 02 = t0 p and t0 < t. Find integer h and i with −s/2 ≤ h ≤ s/2 and
h ≡ u mod s, and −s/2 ≤ i ≤ s/2 and i ≡ v mod s.
Now h2 + i2 = rs for some positive integer r. Note h2 + i2 ≤ (s/2)2 + (s/2)2 < s2 . So r < s. Multiplying this
equation by u2 + v 2 = sp we get (h2 + i2 )(u2 + v 2 ) = rs2 p Note that
(h2 + i2 )(u2 + v 2 ) = (hu + iv)2 + (hv − iu)2 .
Thus
(hu + iv)2 + (hv − iu)2 = rs2 p.
Now since h ≡ u mod s and i ≡ v mod s, hv − iu ≡ 0 mod s. This means that s divides hv − iu, and hence s2 divides
(hu − iu)2 ). By the 2 out of 3 rule (applied to (hu + iv)2 + (hv − iu)2 = rs2 p) we conclude that s divides hu + iv.
Let u0 = (hu + iv)/s and v 0 = (hv − iu)/s. Then u0 and v 0 are integers with u02 + v 02 = rp. Since we’ve already seen
that r < s. we accomplished our task (just set t0 = r).
We can make this kind of substitution only finitely many times, until we end up with p being a sum of squares.
Let’s look at a specific example. Say p = 53. Since p ≡ 1 mod 4, we can find an integer x such that x2 ≡
−1 mod 53. You can check that x = 23 works. In fact 232 + 12 = 530 = 53 · 10. We now find integers u0 and v 0
with −10/2 ≤ u0 , v 0 ≤ 10/2 and u0 ≡ 23 mod 10, and v 0 ≡ 1 mod 10. We can take u0 = 3, and v 0 = 1. Note that
u02 + v 02 = 10 Multiplying the equations 232 + 12 = 530 and u02 + v 02 = 10 together we get
(232 + 12 )(32 + 12 ) = 53 · 10 · 10.
We can now write the LHS as a sum of squares: (23 · 3 + 1 · 1)2 + (23 · 1 − 1 · 3)2 = 702 + 202 = 53 · 10 · 10. Dividing
by 102 we get: 72 + 22 = 53 as desired.
Note sometimes we may not end up with p on the RHS after just one step. In those cases we would have to
continue again.
The final question we consider is: which positive integers can be written as a sum of two squares? Note if
m = a2 + b2 and n = c2 + d2 , then mn = (a2 + b2 )(c2 + d2 ) = (ac + bd)2 − (ad − bc)2 . Thus, if m and n are sums of
2 squares, then mn is the sum of 2 squares.
Since 2 is a sum of 2 squares, and each prime p with p ≡ 1 mod 4 is a sum of 2 squares, whose prime factorization
has no prime q with q ≡ 3 mod 4 to an odd power can be expressed as a sum of two squares. For example,
n = 2 · 32 · 53 · 74 · 116 · 13 is a sum of two squares. One can show the converse is also true: namely if the prime
factorization of n contains a term of the form pe where e is odd, and p ≡ 3 mod 4, then n is not a sum of squares.
Problems
4.1 Use the Law of Quadratic Reciprocity to compute the following:
7
(a) 11
7
11
= − 47 = −1.
11 = − 7
5
(b) 11
5
11
= 15 = 1
11 =
5
2 (c) 2·597·7
2 52 7 2·5 ·7
2
97
=
= 67 = −1 (since the squares mod 7 are 1,2,4).
97
97 · 97 · 97 = 1 · 1 · 7
4.2 Which primes p < 23 are QR mod 23 23 ≡ 7 mod 8, so 2 is a QR mod 23.
is a QR mod
23.
5
23
3
5
2
=
23 =
5
5 = 3 = 3 = −1, so NR.
23
2
7
23 = − 7 = − 7 = −1, so NR.
11
23
2
23 = − 11 = − 11 =−(−1)
= 1, so
is QR.
23
10
5
2
5
13
3
13
=
=
=
·
==
23 13 13 13 13 13 =
5 = 5 = 1, so QR.
17
23
6
2
3
17
2
23 = 17 =
17 =4 17 · 17 = 1 · 3 = 3 = −1 = −1, so NR.
19
23
23 = − 19 = − 19 = −1, so NR.
3
23
=−
23
3
= − 32 = 1, so 3
4.3 Determine which of the following equations have a solution
(a) x2 ≡ 17 mod 23
This is just asking if 17 is a QR mod 23. By the previous problem, the answer is no. So no
solutions.
(b) x2 + 6x ≡ 3 mod 19
Completing
the
square
gives (x + 3)2 ≡ 12 mod 19. So we need to compute
3
1 · 19
= − 19
= − 23 = −1(−1) = 1. So there is a solution.
3
12
19
=
4
19
·
3
19
=
(c) x2 + x ≡ 3 mod 19 (Hint x ≡ 2(10x) mod 19) We complete
with x2 + 20x to get the
the2 square
8
= [ 19 ]3 = (−1)3 = −1, So no solution.
equation: (x + 10)2 ≡ 103 ≡ 8 mod 19. So we look at 19
4.4 Determine which of the following integers are sums of 2 squares:
6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24 Answer: 8, 10, 13,16,17,18, 20
4.5 (a) Write 17 as a sum of two squares , and write 29 as a sum of two squares. 17 = 42 + 12 , and 29 = 52 + 22
(b) Use your answer in (a) to write 17 · 29 as a sum of two squares By the Fermat Formula: 17 · 29 =
(4 · 5 + 1 · 2)2 + (4 · 2 − 1 · 5)2 = (22)2 + 32
(c) Use your answer in (a) to write 52 · 17 · 29 as a sum of two squares. (5 · 22)2 + (5 · 3)2
4.6 Note: Problem has been changed It is known that 932 + 12 = 173 · 50. Use the reduction method to
express 173 as a sum of two squares. We find u and v with −25 ≤ u, v ≤ 25 and u ≡ 93 mod 50 and
v ≡ 1 mod 50. u = −7 and v = 1 work. This gives (−7)2 + 12 = 50 = 50 · 1. Multiplying the
equations together gives: (932 + 12 )((−7)2 + 12 ) = 502 · 173 · 1. Using the Fermat formula we get:
(93 · (−7) + 1 · 1)2 + (93 · 1 − 1 · (−7))2 = 502 · 173 So, 6502 + 1002 = 502 · 173. Dividing by 502 gives:
132 + 22 = 173 · 5.
Dictionary of Definitions of Terms Commonly Used in Matht lectures.
The following is a guide to terms which are commonly used but rarely defined.
CLEARLY: I don’t want to write down all the ”in- between” steps.
TRIVIAL: If I have to show you how to do this, you’re in the wrong class.
OBVIOUSLY: I hope you weren’t sleeping when we discussed this earlier, because I refuse to repeat it.
RECALL: I shouldn’t have to tell you this, but for those of you who erase your memory tapes after every test...
WLOG (Without Loss Of Generality): I’m not about to do all the possible cases, so I’ll do one and let you
figure out the rest.
IT CAN EASILY BE SHOWN: Even you, in your finite wisdom, should be able to prove this without me
holding your hand.
CHECK or CHECK FOR YOURSELF: This is the boring part of the proof, so you can do it on your own
time.
SKETCH OF A PROOF: I couldn’t verify all the details, so I’ll break it down into the parts I couldn’t prove.
HINT: The hardest of several possible ways to do a proof.
BRUTE FORCE (AND IGNORANCE): Four special cases, three counting arguments, two long inductions,
”and a partridge in a pair tree.”
SOFT PROOF: One third less filling (of the page) than your regular proof, but it requires two extra years of
course work just to understand the terms.
ELEGANT PROOF: Requires no previous knowledge of the subject matter and is less than ten lines long.
SIMILARLY: At least one line of the proof of this case is the same as before.
CANONICAL FORM: 4 out of 5 mathematicians surveyed recommended this as the final form for their students
who choose to finish.
TFAE (The Following Are Equivalent): If I say this it means that, and if I say that it means the other thing,
and if I say the other thing...
BY A PREVIOUS THEOREM: I don’t remember how it goes (come to think of it I’m not really sure we did
this at all), but if I stated it right (or at all), then the rest of this follows.
TWO LINE PROOF: I’ll leave out everything but the conclusion, you can’t question ’em if you can’t see ’em.
BRIEFLY: I’m running out of time, so I’ll just write and talk faster.
LET’S TALK THROUGH IT: I don’t want to write it on the board lest I make a mistake.
PROCEED FORMALLY: Manipulate symbols by the rules without any hint of their true meaning (popular in
pure math courses).
QUANTIFY: I can’t find anything wrong with your proof except that it won’t work if x is a moon of Jupiter
(Popular in applied math courses).
PROOF OMITTED: Trust me, It’s true.