Number Theory Problem Set III: Problems & Solutions

Number Theory Problem Set III
Maths Olympiad Preparation
24th February 2024
Problems & Solutions

Find the smallest positive integer m such that for all prime
numbers p > 3, we have
105 | 9p2 − 29p + m.
Since 105 = 3 · 5 · 7, we must have
9p2 − 29p + m ≡ (mod 3, 5, 7).
Since 3 | 9p2 , we have
0 ≡ 29p + m ≡ −(1)p + m ≡ 1 + m (mod 3).
This shows that m ≡ 1 (mod 3). Modulo 5 we also have
0 ≡ 9p2 − 29p + m ≡ (1)p2 − (1)p + m ≡ m (mod 5).
This shows that m ≡ 0 (mod 5). Finally, modulo 7, we have
0 ≡ 9p2 − 29p + m ≡ 2p2 − 1p + m ≡ 2p2 − 1 + m (mod 7).
As p2 ≡ 1 (mod 3) and 23 ≡ 1 (mod 7), we have 2p2 ≡ 2 (mod 7).

Therefore,
0 ≡ 2 − 1 + m = m + 1 (mod 7).
Now, m ≡ −1 (mod 3, 7) is equivalent to m ≡ −1 ≡ 20 (mod 21). Also,
we know that we must have m ≡ 0 ≡ 20 (mod 5). By the Chinese
Remainder Theorem, this is equivalent to m ≡ 20 (mod 105). Hence,
the smallest m is 20.
1
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Determine all triplets (a, b, c) of positive integers that satisfy
a! + bb = c!.
Note that c > a and c > b since c! > bb > b!. Suppose that p is a prime
divisor of b. Then p must divide b!, bb and c!, so that p must divide a!
and p ≤ . Thus, if a = 1, then b = 1 and we get the solution
(a, b, c) = (1, 1, 2).
If a = 2, then b ̸= 1 and the only prime divisor of b is 2. But then bb is

a multiple of 4 and c! ≡ 2 (mod 4). The only possibility is
(a, b, c) = (2, 2, 3).
Suppose, if possible, that a ≥ 3; let q be the largest prime that does

not exceed a. Then, by Bertrand’s postulate that when m ≥ 2 there is
always a prime between m and 2m, a < 2q, so that q 2 cannot divide a!.
However, since q divides a, it must divide c! and hence divide b. Since
a ≥ 3, a! and c! are both even as is b. Because b ̸= 2, we must have that
c ≥ b ≥ 2q (whether q = 2 or q is odd). Hence q 2 divides c! and bb and
so must divide a!, yielding a contradiction.
Therefore, the sole solutions are (a, b, c) = (1, 1, 2), (2, 2, 3).
2 Maths Olympiad Preparation

Prove that for each prime number p > 2, there is exactly one
positive integer n such that the number n2 + np is a perfect
square.
Let n be a positive integer, and p a given prime, p > 2. Then, n2 + np

is a perfect square iff n2 + np = k 2 for some positive integer k, i.e. iff
4n2 + 4np = 4k 2 for some positive integer k. Then,
(2n + p)2 − p2 = 4k 2 ⇐⇒ (2n + p)2 − 4k 2 = p2
i.e. (2n + p − 2k)(2n + p + 2k) = p2 .
Since p is a prime number, the only decompositions of p2 as a product of

two natural numbers are p·p and 1·p2 . As k > 0, 2n+p+2k > 2n+p−2k,
so it cannot be
2n + p + 2k = 2n + p − 2k = p.
Therefore, it must be
2n + p + 2k = p2 , 2n + p − 2k = 1.
Adding them up:

2
p2 − 2p + 1

2 p−1
4n + 2p = p + 1 ⇐⇒ n = = .
4 2
2
p−1
So n = is the only value that makes n2 +np a perfect square.
2
Notice that p > 2 is prime, so it is an odd number, so p − 1 is even, which
implies that n is a positive integer.
QED.
If a, b, c & d ∈ N such that gcd(a, b, c, d) = 1 and abcd | a2 +b2 +c2 +d2 .

Prove that
a2 + b2 + c2 + d2
= 4.
abcd

Consider the equation a2 + b2 + c2 + d2 = kabcd for a fixed k ∈ N, and

assume that (a, b, c, d) is a solution with minimal sum a + b + c + d. We
can also assume wlog that a ≥ b ≥ c ≥ d. Now the second root a1 of the
quadratic
f (x) = x2 − kbcdx + b2 + c2 + d2
satisfies by Vieta
b2 + c2 + d2
a1 = = kbcd − a,
a
hence it is also a positive integer. Since (a1 , b, c, d) is also a solution,
we must have a1 ≥ a. In particular, a2 ≤ b2 + c2 + d2 (∗). Moreover,
since b is outside of the interval [a, a1 ] between the roots of the quadratic
function f , we have f (b) ≥ 0. On the other hand,
f (b) ≤ 4b2 − kb2 cd = b2 (4 − kcd) =⇒ kcd ≤ 4.
This already shows that k ∈ {1, 2, 3, 4}, and we need to prove that k =
1, 2, 3 do not work. Since also cd ≤ 4, this can be done by considering
the few possible cases by hand:
Case I: If c = d = 1 then from (∗) we have a2 ≤ b2 + 2 < (b + 1)2 =⇒
a = b. Hence, a2 | 2(a2 + 2), which is only possible for a = 1. This gives
the solution a = b = c = d = 1, k = 4.
Case II: If c = d = 2 then k = 1 and we get a2 + b2 + 8 = 4ab. Con-
sidering this modulo 4 we get 2 | a, b which contradicts the assumption
(in fact in this case the unique solution is a = b = 2).
Case III: If c = 2, d = 1 we have k ∈ {1, 2} and a2 + b2 + 5 = 2ab or
a2 + b2 + 5 = 4ab. The first one is clearly impossible. In the second case,
from (∗) we have a2 = b2 + 5 < (b + 2)2 . Since a + b must be odd this
yields a = b + 1, which does not lead to a solution.
Case IV: If c = 3, d = 1 we have k = 1 =⇒ a2 + b2 + 10 = 3ab. Also,
(∗) implies a2 = b2 + 10 < (b + 3)2 . Furthermore a, b must have the same
parity and thus a ∈ {b, b + 2}, both of which can easily be seen not to
work.
Case V: If c = 4, d = 1 then k = 1 =⇒ a2 + b2 + 17 = 4ab. Also, (∗)
implies a2 ≤ b2 + 17 < (b + 2)2 , since b ≥ 4. Hence a = b + 1, which does
not lead to a solution.
This also shows that there are only two minimal solutions: (1, 1, 1, 1)
with k = 4, and (2, 2, 2, 2) with k = 1.
QED.

Solve in positive integers:
x4 = y 6 + y 4 + 1.
q
Let x2 = 2 and rewrite the original equation as
q 2 = (2y 3 + y)2 + 4 − y 2 .
But for y > 2, we have
(2y 3 + y − 1)2 < (2y 3 + y)2 + 4 − y 2 < (2y 3 + y)2 .
Thus
2y 3 + y − 1 < q < 2y 3 + y.
But this is not possible.
Hence y = 1, 2. Checking gives (x, y) = (3, 2).
Thank You All!

Number Theory Problem Set III: Problems & Solutions

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Number Theory Problem Set III

Maths Olympiad Preparation

24th February 2024

Problems & Solutions

105 | 9p2 − 29p + m.

Since 105 = 3 · 5 · 7, we must have

9p2 − 29p + m ≡ (mod 3, 5, 7).

Since 3 | 9p2 , we have

0 ≡ 29p + m ≡ −(1)p + m ≡ 1 + m (mod 3).

This shows that m ≡ 1 (mod 3). Modulo 5 we also have

0 ≡ 9p2 − 29p + m ≡ (1)p2 − (1)p + m ≡ m (mod 5).

This shows that m ≡ 0 (mod 5). Finally, modulo 7, we have

0 ≡ 9p2 − 29p + m ≡ 2p2 − 1p + m ≡ 2p2 − 1 + m (mod 7).

As p2 ≡ 1 (mod 3) and 23 ≡ 1 (mod 7), we have 2p2 ≡ 2 (mod 7).

Determine all triplets (a, b, c) of positive integers that satisfy

(a, b, c) = (1, 1, 2).

If a = 2, then b ̸= 1 and the only prime divisor of b is 2. But then bb is

(a, b, c) = (2, 2, 3).

Suppose, if possible, that a ≥ 3; let q be the largest prime that does

2 Maths Olympiad Preparation

Let n be a positive integer, and p a given prime, p > 2. Then, n2 + np

(2n + p)2 − p2 = 4k 2 ⇐⇒ (2n + p)2 − 4k 2 = p2

i.e. (2n + p − 2k)(2n + p + 2k) = p2 .

Since p is a prime number, the only decompositions of p2 as a product of

Adding them up:

If a, b, c & d ∈ N such that gcd(a, b, c, d) = 1 and abcd | a2 +b2 +c2 +d2 .

3 Maths Olympiad Preparation

Consider the equation a2 + b2 + c2 + d2 = kabcd for a fixed k ∈ N, and

f (b) ≤ 4b2 − kb2 cd = b2 (4 − kcd) =⇒ kcd ≤ 4.

4 Maths Olympiad Preparation

Solve in positive integers:

But for y > 2, we have

(2y 3 + y − 1)2 < (2y 3 + y)2 + 4 − y 2 < (2y 3 + y)2 .

Thank You All!

5 Maths Olympiad Preparation

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