# Questions based on Wilson’s theorem for training for RMO

1(a) Find the remainder when $15!$ is divided by 17.
1(b) Find the remainder when $2(26!)$ is divided by 29.

2: Determine whether 17 is a prime by deciding if $16! \equiv -1 {\pmod 17}$

3: Arrange the integers 2,3,4, …, 21 in pairs a and b that satisfy $ab \equiv 1 {\pmod 23}$.

4: Show that $18! \equiv -1 {\pmod 437}$.

5a: Prove that an integer $n>1$ is prime if and only if $(n-2)! \equiv 1 {\pmod n}$.
5b: If n is a composite integer, show that $(n-1)! \equiv 0 {\pmod n}$, except when $n=4$.

6: Given a prime number p, establish the congruence $(p-1)! \equiv {p-1} {\pmod {1+2+3+\ldots + (p-1)}}$

7: If p is prime, prove that for any integer a, $p|a^{p}+(p-1)|a$ and $p|(p-1)!a^{p}+a$

8: Find two odd primes $p \leq 13$ for which the congruence $(p-1)! \equiv -1 {\pmod p^{2}}$ holds.

9: Using Wilson’s theorem, prove that for any odd prime p:
$1^{2}.3^{2}.5^{2}.\ldots (p-2)^{2} \equiv (-1)^{(p+1)/2} {\pmod p}$

10a: For a prime p of the form $4k+3$, prove that either

$(\frac{p-1}{2})! \equiv 1 {\pmod p}$ or $(\frac{p-1}{2})! \equiv -1 {\pmod p}$

10b: Use the part (a) to show that if $4k+3$ is prime, then the product of all the even integers less than p is congruent modulo p to either 1 or -1.

More later,
Nalin Pithwa.

# Wilson’s theorem and related problems in Elementary Number Theory for RMO

I) Prove Wilson’s Theorem:

If p is a prime, then $(p-1)! \equiv -1 {\pmod p}$.

Proof:

The cases for primes 2 and 3 are clearly true.

Assume $p>3$

Suppose that a is any one of the p-1 positive integers $1,2,3, \ldots {p-1}$ and consider the linear congruence
$ax \equiv 1 {\pmod p}$. Then, $gcd(a,p)=1$.

Now, apply the following theorem: the linear congruence $ax \equiv b {\pmod n}$ has a solution if and only if $d|b$, where $d = gcd(a,b)$. If $d|b$, then it has d mutually incongruent solutions modulo n.

So, by the above theorem, the congruence here admits a unique solution modulo p; hence, there is a unique integer $a^{'}$, with $1 \leq a^{'} \leq p-1$, satisfying $aa^{'} \equiv 1 {\pmod p}$.

Because p is prime, $a = a^{'}$ if and only if $a=1$ or $a=p-1$. Indeed, the congruence $a^{2} \equiv 1 {\pmod p}$ is equivalent to $(a-1)(a+1) \equiv 0 {\pmod p}$. Therefore, either $a-1 \equiv 0 {\pmod p}$, in which case $a=1$, or $a+1 \equiv 0 {\pmod p}$, in which case $a=p-1$.

If we omit the numbers 1 and p-1, the effect is to group the remaining integers $2,3, \ldots (p-2)$ into pairs $a$ and $a^{'}$, where $a \neq a^{'}$, such that the product $aa^{'} \equiv 1 {\pmod p}$. When these $(p-3)/2$ congruences are multiplied together and the factors rearranged, we get

$2.3. \ldots (p-2) \equiv 1 {\pmod p}$

or rather

$(p-2)! \equiv 1 {\pmod p}$

Now multiply by p-1 to obtain the congruence

$(p-1)! \equiv p-1 \equiv -1 {\pmod p}$, which was desired to be proved.

An example to clarify the proof of Wilson’s theorem:

Specifically, let us take prime $p=13$. It is possible to divide the integers $2,3,4, \ldots, 11$ into $(p-3)/2=5$ pairs, each product of which is congruent to 1 modulo 13. Let us write out these congruences explicitly as shown below:

$2.7 \equiv 1 {\pmod {13}}$
$3.9 \equiv 1 {\pmod {13}}$
$4.10 \equiv 1 {\pmod {13}}$
$5.8 \equiv 1 {\pmod {13}}$
$6.11 \equiv 1 {\pmod {13}}$

Multpilying these congruences gives the result $11! = (2.7)(3.9)(4.10)(5.8)(6.11) \equiv 1 {\pmod {13}}$

and as $12! \equiv 12 \equiv -1 {\pmod {13}}$

Thus, $(p-1)! \equiv -1 {\pmod p}$ with prime $p=13$.

Further:

The converse to Wilson’s theorem is also true. If $(n-1)! \equiv -1 {\pmod n}$, then n must be prime. For, if n is not a prime, then n has a divisor d with $1 1$ is prime if and only if $(n-1)! \equiv -1 {\pmod n}$. Unfortunately, this test is of more theoretical than practical interest because as n increases, $(n-1)!$ rapidly becomes unmanageable in size.

Let us illustrate an application of Wilson’s theorem to the study of quadratic congruences{ What we mean by quadratic congruence is a congruence of the form $ax^{2}+bx+c \equiv 0 {\pmod n}$, with $a \not\equiv 0 {\pmod n}$ }

Theorem: The quadratic congruence $x^{2}+1 \equiv 0 {\pmod p}$, where p is an odd prime, has a solution if and only if $p \equiv 1 {\mod 4}$.

Proof:

Let a be any solution of $x^{2}+1 \equiv 0 {\pmod p}$ so that $a^{2} \equiv -1 {\pmod p}$. Because $p \not |a$, the outcome of applying Fermat’s Little Theorem is

$1 \equiv a^{p-1} \equiv (a^{2})^{(p-1)/2} \equiv (-1)^{(p-1)/2} {\pmod p}$

The possibility that $p=4k+3$ for some k does not arise. If it did, we would have

$(-1)^{(p-1)/2} = (-1)^{2k+1} = -1$

Hence, $1 \equiv -1 {\pmod p}$. The net result of this is that $p|2$, which is clearly false. Therefore, p must be of the form $4k+1$.

Now, for the opposite direction. In the product

$(p-1)! = 1.2 \ldots \frac{p-1}{2} \frac{p+1}{2} \ldots (p-2)(p-1)$

we have the congruences

$p-1 \equiv -1 {\pmod p}$
$p-2 \equiv -2 {\pmod p}$
$p-3 \equiv -3 {\pmod p}$
$\vdots$
$\frac{p+1}{2} \equiv - \frac{p-1}{2} {\pmod p}$

Rearranging the factors produces
$(p-1)! \equiv 1.(-1).2.(-2) \ldots \frac{p-1}{2}. (-\frac{p-1}{2}) {\pmod p} \equiv (-1)^{(p-1)/2}(.2. \ldots \frac{p-1}{2})^{2}{\pmod p}$

because there are $(p-1)/2$ minus signs involved. It is at this point that Wilson’s theorem can be brought to bear; for, $(p-1)! \equiv -1 {\pmod p}$, hence,

$-1 \equiv (-1)^{(p-1)/2}((\frac{p-1}{2})!)^{2} {\pmod p}$

If we assume that p is of the form $4k+1$, then $(-1)^{(p-1)/2} =1$, leaving us with the congruence

$-1 \equiv (-\frac{p-1}{2})^{2}{\pmod p}$.

The conclusion is that the integer $(\frac{p-1}{2})!$ satisfies the quadratic congruence $x^{2}+1 \equiv 0 {\pmod p}$.

Let us take a look at an actual example, say, the case $p=13$, which is a prime of the form $4k+1$. Here, we have $\frac{p-1}{2}=6$, and it is easy to see that $6! = 720 \equiv 5 {\pmod {13}}$ and $5^{2}+1 = 26 \equiv 0 {\pmod {13}}$.

Thus, the assertion that $((p-1)!)^{2}+1 \equiv 0 {\pmod p}$ is correct for $p=13$.

Wilson’s theorem implies that there exists an infinitude of composite numbers of the form $n!+1$. On the other hand, it is an open question whether $n!+1$ is prime for infinitely many values of n. Refer, for example:

https://math.stackexchange.com/questions/949520/are-there-infinitely-many-primes-of-the-form-n1

More later! Happy churnings of number theory!
Regards,
Nalin Pithwa

# Eight digit bank identification number and other problems of elementary number theory

Question 1:

Consider the eight-digit bank identification number $a_{1}a_{2}\ldots a_{8}$, which is followed by a ninth check digit $a_{9}$ chosen to satisfy the congruence

$a_{9} \equiv 7a_{1} + 3a_{2} + 9a_{3} + 7a_{4} + 3a_{5} + 9a_{6} + 7a_{7} + 3a_{8} {\pmod {10}}$

(a) Obtain the check digits that should be appended to the two numbers 55382006 and 81372439.

(b) The bank identification number $237a_{4}18538$ has an illegitimate fourth digit. Determine the value of the obscured digit.

Question 2:

(a) Find an integer having the remainders 1,2,5,5 when divided by 2, 3, 6, 12 respectively (Yih-hing, died 717)

(b) Find an integer having the remainders 2,3,4,5 when divided by 3,4,5,6 respectively (Bhaskara, born 1114)

(c) Find an integer having remainders 3,11,15 when divided by 10, 13, 17, respectively (Regiomontanus, 1436-1476)

Question 3:

Question 3:

Let $t_{n}$ denote the nth triangular number. For which values of n does $t_{n}$ divide $t_{1}^{2} + t_{2}^{2} + \ldots + t_{n}^{2}$

Hint: Because $t_{1}^{2}+t_{2}^{2}+ \ldots + t_{n}^{2} = t_{n}(3n^{3}+12n^{2}+13n+2)/30$, it suffices to determine those n satisfying $3n^{3}+12n^{2}+13n+2 \equiv 0 {\pmod {2.3.5}}$

Question 4:

Find the solutions of the system of congruences:

$3x + 4y \equiv 5 {\pmod {13}}$
$2x + 5y \equiv 7 {\pmod {13}}$

Question 5:

Obtain the two incongruent solutions modulo 210 of the system

$2x \equiv 3 {\pmod 5}$
$4x \equiv 2 {\pmod 6}$
$3x \equiv 2 {\pmod 7}$

Question 6:

Use Fermat’s Little Theorem to verify that 17 divides $11^{104}+1$

Question 7:

(a) If $gcd(a,35)=1$, show that $a^{12} \equiv {\pmod {35}}$. Hint: From Fermat’s Little Theorem, $a^{6} \equiv 1 {\pmod 7}$ and $a^{4} \equiv 1 {\pmod 5}$

(b) If $gcd(a,42) =1$, show that $168=3.7.8$ divides $a^{6}-1$
(c) If $gcd(a,133)=gcd(b,133)=1$, show that $133| a^{18} - b^{18}$

Question 8:

Show that $561|2^{561}-1$ and $561|3^{561}-3$. Do there exist infinitely many composite numbers n with the property that $n|2^{n}-2$ and $n|3^{n}-3$?

Question 9:

Prove that any integer of the form $n = (6k+1)(12k+1)(18k+1)$ is an absolute pseudoprime if all three factors are prime; hence, $1729=7.13.19$ is an absolute pseudoprime.

Question 10:

Prove that the quadratic congruence $x^{2}+1 \equiv 0 {\pmod p}$, where p is an odd prime, has a solution if and only if $p \equiv {pmod 4}$.

Note: By quadratic congruence is meant a congruence of the form $ax^{2}+bx+c \equiv 0 {\pmod n}$ with $a \equiv 0 {\pmod n}$. This is the content of the above proof.

More later,
Nalin Pithwa.

# Pre RMO algebra : some tough problems

Question 1:

Find the cube root of $x^{3} -12x^{2} + 54x -112 + \frac{108}{x} - \frac{48}{x^{2}} + \frac{8}{x^{3}}$

Question 2:

Find the square root of $\frac{x}{y} + \frac{y}{x} +3 - 2\sqrt{\frac{x}{y}} -2\sqrt{\frac{y}{x}}$

Question 3:

Simplify (a):

$(\frac{x}{x-1} - \frac{1}{x+1}). \frac{x^{3}-1}{x^{6}+1}.\frac{(x-1)^{2}(x+1)^{2}+x^{2}}{x^{4}+x^{2}+1}$

Simplify (b):
$\{ \frac{a^{4}-y^{4}}{a^{2}-2ay+y^{2}} \div \frac{a^{2}+ay}{a-y} \} \times \{ \frac{a^{5}-a^{3}y^{2}}{a^{3}+y^{3}} \div \frac{a^{4}-2a^{3}y+a^{2}y^{2}}{a^{2}-ay+y^{2}}\}$

Question 4:

Solve : $\frac{3x}{11} + \frac{25}{x+4} = \frac{1}{3} (x+5)$

Question 5:

Solve the following simultaneous equations:

$2x^{2}-3y^{2}=23$ and $2xy - 3y^{2}=3$

Question 6:

Simplify (a):

$\frac{1- \frac{a^{2}}{(x+a)^{2}}}{(x+a)(x-a)} \div \frac{x(x+2a)}{(x^{2}-a^{2})(x+a)^{2}}$

Simplify (b):

$\frac{6x^{2}y^{2}}{m+n} \div \{\frac{3(m-n)x}{7(r+s)} \div \{ \frac{4(r-s)}{21xy^{2}} \div \frac{(r^{2}-s^{2})}{4(m^{2}-n^{2})}\} \}$

Question 7:

Find the HCF and LCM of the following algebraic expressions:

$20x^{4}+x^{2}-1$ and $25x^{4}+5x^{3} - x - 1$ and $25x^{4} -10x^{2} +1$

Question 8:

Simplify the following using two different approaches:

$\frac{5}{6- \frac{5}{6- \frac{5}{6-x}}} = x$

Question 9:

Solve the following simultaneous equations:

Slatex x^{2}y^{2} + 192 = 28xy\$ and $x+y=8$

Question 10:

If a, b, c are in HP, then show that

$(\frac{3}{a} + \frac{3}{b} - \frac{2}{c})(\frac{3}{c} + \frac{3}{b} - \frac{2}{a})+ \frac{9}{b^{2}}=\frac{25}{ac}$

Question 11:

if $a+b+c+d=2s$, prove that

$4(ab+cd)^{2} - (a^{2}+b^{2}-c^{2}-d^{2})^{2}= 16(s-a)(s-b)(s-c)(s-d)$

Question 12:

Determine the ratio $x:y:z$ if we know that

$\frac{x+z}{y} = \frac{z}{x} = \frac{x}{z-y}$

More later,
Nalin Pithwa

Those interested in such mathematical olympiads should refer to: