nyuuzyou/wikis huggingface dataset:
Number whose Square and Cube use all Digits Once

== Theorem ==
The only integer whose square and cube use each of the digits from $0$ to $9$ exactly once each is $69$.
== Theorem ==
First the limits are established for the square and cube of an integer to have exactly $10$ digits between them.
So integers smaller than $47$ do not have enough digits available, and integers greater than $99$ have too many.
$47$ and $99$ themselves are eliminated on inspection of the above.
It is noted that integers ending in $0$, $1$, $5$ and $6$ have squares and cubes ending in those same digits.
Such numbers can be eliminated from our search, as they will duplicate the appearance of those digits.
It remains to check the integers between $48$ and $98$.
First the squares which duplicate at least one digit are eliminated:
Of the remainder, the cubes which duplicate at least one digit are eliminated:
Of the remainder, the squares and cubes which duplicate at least one digit between them are eliminated:
Finally:
The result follows.
== Sources ==
* : $69$
* : $69$
Category:Recreational Mathematics
Category:Square Numbers
Category:Cube Numbers
Category:69
Number whose Square is in 2 Identical Halves

== Theorem ==
The number $36 \, 363 \, 636 \, 364$ has the property that its square can be split into two identical halves:
:$36 \, 363 \, 636 \, 364 = 1 \, 322 \, 314 \, 049 \, 613 \, 223 \, 140 \, 496$
== Proof ==
It just does.
== Historical Note ==
== Sources ==
* : $36,363,636,364$
* : $36,363,636,364$
Category:Square Numbers
Category:36,363,636,364
Numbers Appearing 8 Times in Pascal's Triangle

== Theorem ==
Excluding $1$, the number $3003$ is the smallest integer to appear $8$ times in Pascal's triangle.
No other number below $2^{23}$ appears as often.
== Proof ==
== Also see ==
* Singmaster's Conjecture
== Sources ==
* 
* : $3003$
* : $3003$
Category:Pascal's Triangle
Category:3003
Numbers Divisible by Sum and Product of Digits

== Theorem ==
The sequence of positive integers which are divisible by both the sum and product of its digits begins:
:$1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 24, 36, 111, 112, 132, 135, \ldots$
== Proof ==
Let $S$ be the set of all positive integers which are divisible by both the sum and product of their digits.
Trivially, the sum and product of the digits of a one-digit number $n$ are themselves $n$.
Thus from Integer Divides Itself, the positive integers from $1$ to $9$ are in $S$.
The product of any integer with a $0$ in it is $0$.
From Zero Divides Zero $0$ is not a divisor of $n$ unless $n$ is itself $0$.
So $10, 20, 30, \ldots$ are not in $S$.
For all prime numbers $p$ with $2$ or more digits, the sum of its digits is greater than $1$ and less than $p$.
Thus $p$ is not a multiple of the sum of its digits.
== Sources ==
* 
::This citation is given in , but it is doubtful.
* : $24$
Category:Number Theory
Numbers Equal to Number of Digits in Factorial

== Theorem ==
For $n \in \set {1, 22, 23, 24}$ the number of digits in the decimal representation of $n!$ is equal to $n$.
== Proof ==
First we note that:
:$1! = 1$
which has $1$ digit.
Then from Examples of Factorials:
:$2! = 2$
which has $1$ digit.
Multiplying $n$ by a $1$-digit number increases the number of digits of $n$ by no more than $1$.
Thus from $n = 3$ to $n = 9$, $n!$ has no more than $n - 1$ digits.
From $21$ Factorial:
:$21! = 51 \, 090 \, 942 \, 171 \, 709 \, 440 \, 000$
and so $21!$ has $20$ digits.
Multiplying $n$ by a $2$-digit number increases the number of digits of $n$ by at least $1$.
Thus from $n = 10$ to $n = 20$, $n!$ has no more than $n - 1$ digits.
It is noted that:
:From $22$ Factorial:
::$22! = 1 \, 124 \, 000 \, 727 \, 777 \, 607 \, 680 \, 000$
:which has $22$ digits.
:From $23$ Factorial:
::$23! = 25 \, 852 \, 016 \, 738 \, 884 \, 976 \, 640 \, 000$
:which has $23$ digits.
:From $24$ Factorial:
::$24! = 620 \, 448 \, 401 \, 733 \, 239 \, 439 \, 360 \, 000$
:which has $24$ digits.
But from $25$ Factorial:
:$25! = 15 \, 511 \, 210 \, 043 \, 330 \, 985 \, 984 \, 000 \, 000$
which has $26$ digits.
As above, multiplying $n$ by a number with $2$ digits or more increases the number of digits of $n$ by at least $1$.
So if $n!$ has more than $n$ digits, $\paren {n + 1}!$ has more than $n + 1$ digits.
It follows by induction that for all $n \ge 25$, $n!$ has more than $n$ digits.
Hence the result.
== Sources ==
* : $22$
* : $22$
Category:Factorials
Numbers Equal to Sum of Squares of Digits

== Theorem ==
There are exactly $2$ integers which are equal to the sum of the squares of their digits when expressed in base $10$:
:$0 = 0^2$
:$1 = 1^2$
== Proof ==
We see the cases $0$ and $1$ above hold.
Suppose $N > 1$ is equal to the sum of the squares of their digits when expressed in base $10$.
Since $N^2 > N$, $N$ cannot be a $1$-digit integer.
Suppose $N$ is a $2$-digit integer.
Write $N = \sqbrk {a b} = 10 a + b$.
Then we have $a^2 + b^2 = 10 a + b$.
This can be reduced to $b \paren {b - 1} = a \paren {10 - a}$.
Since $b \paren {b - 1}$ is even, $a$ must be even as well.
The list of possible values of $a \paren {10 - a}$ are:
:$2 \paren {10 - 2} = 8 \paren {10 - 8} = 16$
:$4 \paren {10 - 4} = 6 \paren {10 - 6} = 24$
The list of possible values of $b \paren {b - 1}$ are:
:$b \paren {b - 1} \le 4 \paren {4 - 1} = 12$ for $b \le 4$
:$5 \paren {5 - 1} = 20$
:$b \paren {b - 1} \ge 6 \paren {6 - 1} = 30$ for $b \ge 6$
We see that they do not coincide.
Thus $N$ cannot be a $2$-digit integer.
Suppose $100 \le N \le 199$.
Write $N = \sqbrk {1 a b} = 100 + 10 a + b$.
Then we have $1^2 + a^2 + b^2 = 100 + 10 a + b$.
This can be reduced to $b \paren {b - 1} = a \paren {10 - a} + 99$.
But we have $b \paren {b - 1} < 9 \times 8 = 72 < 99 \le a \paren {10 - a} + 99$.
So $N$ cannot be in this range.
Suppose $200 \le N \le 299$.
Then the sum of the squares of their digits cannot exceed $2^2 + 9^2 + 9^2 = 186 < 200$.
So $N$ cannot be in this range.
Suppose $300 \le N \le 999$.
Then the sum of the squares of their digits cannot exceed $9^2 + 9^2 + 9^2 = 243 < 300$.
So $N$ cannot be in this range.
Suppose $N$ is a $k$-digit integer with $k \ge 4$.
Then the sum of the squares of their digits cannot exceed $9^2 \times k$.
We have:
So $N$ cannot be greater than $1$.
== Historical Note ==
== Sources ==
* : $0$
Category:Number Theory
Numbers Equal to Sum of Squares of Digits/Historical Note

== Historical Note on Numbers Equal to Sum of Squares of Digits ==
This result was presented without proof in of $1983$.
However, they failed to note that it is important to stress the fact that it depends on the number base.
While it holds for base $10$, it is not true for base $3$:
:$5$ expressed in base $3$ is $12$, while $1^2 + 2^2 = 5$
Category:Historical Notes
Numbers Expressed as Sums of Binomial Coefficients

== Definition ==
Let $n \in \Z_{> 0}$ be a (strictly) positive integer.
Then for all $k \in \Z_{> 0}$, it is possible to express $k$ uniquely in the form:
such that $0 \le k_1 < k_2 < \cdots < k_n$.
== Proof ==
=== Existence of Representation ===
Proof by induction:
For all $k \in \Z_{\ge 0}$, let $\map P k$ be the proposition that it is possible to express $k$ in the form:
:$\ds k = \sum_{j \mathop = 1}^n \dbinom {k_j} j$
such that $0 \le k_1 < k_2 < \cdots < k_n$.
==== Basis for the Induction ====
$\map P 0$ is true, as $\ds 0 = \sum_{j \mathop = 1}^n \dbinom {j - 1} j$.
This is our basis for the induction.
==== Induction Hypothesis ====
Now we need to show that, if $\map P m$ is true, where $r \ge 2$, then it logically follows that $\map P {m + 1}$ is true.
So this is our induction hypothesis:
$m$ can be expressed in the form:
:$\ds m = \sum_{j \mathop = 1}^n \dbinom {m_j} j$
where $0 \le m_1 < m_2 < \cdots < m_n$.
Then we need to show:
$m + 1$ can be expressed in the form:
:$\ds m = \sum_{j \mathop = 1}^n \dbinom {k_j} j$
where $0 \le k_1 < k_2 < \cdots < k_n$.
==== Induction Step ====
This is our induction step:
Suppose the first $c$ $m_j$ are consecutive, that is:
:$\forall \, j \in \N: 1 \le j < c: m_{j + 1} - m_j = 1$ and $m_{c + 1} - m_c > 1$
Then:
Since $0 \le m_1 < m_2 < \cdots < m_n$, we must have $m_j \ge j - 1$ for each $j \le n$.
Hence $0 \le 0 < 1 < \cdots < c - 2 < m_c - 1 < m_{c + 1} < \cdots < m_n$.
Thus the expression above satisfy the conditions.
So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \Z_{\ge 0}: \sum_{k \mathop = 0}^n \binom {r + k} k = \binom {r + n + 1} n$
=== Uniqueness of Representation ===
Suppose $k$ can be expressed in the form:
:$\ds k = \sum_{j \mathop = 1}^n \dbinom {k_j} j = \sum_{j \mathop = 1}^n \dbinom {m_j} j$
where $0 \le k_1 < k_2 < \cdots < k_n$ and $0 \le m_1 < m_2 < \cdots < m_n$.
 $k_i$ and $m_i$ are not all equal.
Let $c$ be the largest integer such that $k_c \ne m_c$.
 assume $k_c < m_c$.
Then:
which is a contradiction.
Thus the representation is unique.
== Examples ==
== Also see ==
* Definition:Combinatorial Number System
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $56$
Category:Numbers Expressed as Sums of Binomial Coefficients
Category:Binomial Coefficients
Numbers Expressible as Sum of Five Distinct Squares

== Theorem ==
The largest positive integer which cannot be expressed as the sum of no more than $5$ distinct squares is $188$.
Both $188$ and $124$ require as many as $6$ distinct squares to represent them:
== Proof ==
From Numbers not Sum of Distinct Squares, the following positive integers cannot be expressed as the sum of distinct squares at all:
:$2, 3, 6, 7, 8, 11, 12, 15, 18, 19, 22, 23, 24, 27, 28, 31, 32, 33, 43, 44, 47, 48, 60, 67, 72, 76, 92, 96, 108, 112, 128$
== Sources ==
* 
* 
* : $188$
Category:Specific Numbers
Category:188
Numbers Not Expressible as Sum of no more than 5 Squares of Composite Numbers

== Theorem ==
There are $256$ integers which cannot be expressed as the sum of no more than $5$ squares of composite numbers:
:$1, 2, 3, \ldots, 1167$
== Sources ==
* : $\mathbf C$: Additive Number Theory: $\mathbf {C 20}$: Sums of Squares
* : $1167$
Category:Number Theory
Numbers Not Expressible as Sum of no more than 5 Squares of Composite Numbers/Mistake

== Source Work ==
:
:The Dictionary
::$1167$
== Mistake ==
:[$1167$ is] The largest of $256$ numbers which cannot be expressed as the sum of at most $5$ composite numbers. [Guy 136]
== Correction ==
This should say:
:[$1167$ is] The largest of $256$ numbers which cannot be expressed as the sum of at most $5$ squares of composite numbers. [Guy 136]
(emphasis added)
The mistake appears in (and has been copied from) 's of $1994$.
However, in this has been corrected.
== Sources ==
* : $\mathbf C$: Additive Number Theory: $\mathbf {C 20}$: Sums of Squares
* : $1167$
Category:Mistakes/Curious and Interesting Numbers
Numbers Reversed when Multiplying by 4

== Theorem ==
Numbers of the form $\sqbrk {21 \paren 9 78}_{10}$ are reversed when they are multiplied by $4$:
and so on.
== Proof ==
Let k represent the number of $9$s in the middle of the number.
For $k > 0$ We can rewrite the number as follows:
Taking numbers of this form and multiplying by $4$ produces:
The first part is composed of $k + 4$ digits. The first two digits will be $84$ followed by $k +2$ digits of $0$
The sum in the middle is composed of $k + 3$ digits. The first digit will be $3$ followed by $k - 1$ digits of $9$ and then the remaining three digits at the end are $600$
Summing the three pieces, the final answer will have $k + 4$ digits.
The first digit is $8$ 
followed by $7$ which is the sum of the $4$ from the first part and the $3$ of the middle part
followed by $k$ digits of $9$ where the last $9$ is the sum of the $6$ from the middle part and the $3$ of the last part
and then ending in $12$:
== Also see ==
* Numbers Reversed when Multiplying by 9
== Sources ==
* : $2178$
* : $2178$
Category:Reversals
Category:2178
Numbers Reversed when Multiplying by 9

== Theorem ==
Numbers of the form $\sqbrk {10 (9) 89}_{10}$ are reversed when they are multiplied by $9$:
and so on.
== Proof ==
Let k represent the number of $9$s in the middle of the number.
For $k > 0$ We can rewrite the number as follows:
Taking numbers of this form and multiplying by $9$ produces:
The first part is composed of $k + 4$ digits. The first digit will be $9$ followed by $k + 3$ digits of $0$
The sum in the middle is composed of $k + 3$ digits. The first digit will be $8$ followed by $k - 1$ digits of $9$ and then the remaining three digits at the end are $100$
Summing the three pieces, the final answer will have $k + 4$ digits.
The first digit is $9$ 
followed by $8$ which is the first digit of the middle part
followed by $k$ digits of $9$ where the last $9$ is the sum of the $1$ from the middle part and the $8$ of the last part
and then ending in $01$:
== Also see ==
* Numbers Reversed when Multiplying by 4
== Sources ==
* : $1089$
* : $1089$
Category:Reversals
Category:1089
Numbers are Coprime iff Sum is Coprime to Both

== Theorem ==
Let $a, b$ be integers.
Then:
:$a \perp b \iff a \perp \paren {a + b}$
where $a \perp b$ denotes that $a$ and $b$ are coprime.
== Proof ==
=== Necessary Condition ===
Let $a \perp b$.
Suppose $a + b$ is not coprime to $a$.
Then:
:$\exists d \in \Z_{>1}: d \divides a, d \divides \paren {a + b}$
But then:
:$d \divides \paren {\paren {a + b} - a}$
and so:
:$d \divides b$
and so $a$ and $b$ are not coprime.
From this contradiction it follows that $a + b$ is coprime to $a$.
=== Sufficient Condition ===
Let $a + b$ be coprime to $a$.
Suppose $a$ is not coprime to $b$.
Then:
:$\exists d \in \Z_{>1}: d \divides a, d \divides b$
and so:
:$d \divides \paren {a + b}$
and so $a$ and $\paren {a + b}$ are not coprime.
From this contradiction it follows that $a$ is coprime to $b$.
== Sources ==
* : Book $\text{VII}$. Propositions
Category:Coprime Integers
Numbers between which exist two Mean Proportionals are Similar Solid

== Theorem ==
Let $a, b \in \Z$ be the extremes of a geometric sequence of integers whose length is $4$:
:$\tuple {a, m_1, m_2, b}$
That is, such that $a$ and $b$ have $2$ mean proportionals.
Then $a$ and $b$ are similar solid numbers.
== Proof ==
From Form of Geometric Sequence of Integers:
:$\exists k, p, q \in \Z: a = k p^3, b = k q^3$
So $a$ and $b$ are solid numbers whose sides are:
:$k p$, $p$ and $p$
and
:$k q$, $q$ and $q$
respectively.
Then:
:$\dfrac {k p} {k q} = \dfrac p q$
demonstrating that $a$ and $b$ are similar solid numbers by definition.
== Sources ==
* : Book $\text{VIII}$. Propositions
Category:Euclidean Number Theory
Numbers between which exists one Mean Proportional are Similar Plane

== Theorem ==
Let $a, b \in \Z$ such that the geometric mean is an integer.
Then $a$ and $b$ are similar plane numbers.
== Proof ==
Let the geometric mean of $a$ and $b$ be an integer $m$.
Then, in the language of , $m$ is a mean proportional of $a$ and $b$.
Thus:
:$\left({a, m, b}\right)$
is a geometric sequence.
From Form of Geometric Sequence of Integers:
:$\exists k, p, q \in \Z: a = k p^2, b = k q^2$
So $a$ and $b$ are plane numbers whose sides are:
:$k p$ and $p$
and
:$k q$ and $q$
respectively.
Then:
:$\dfrac {k p} {k q} = \dfrac p q$
demonstrating that $a$ and $b$ are similar plane numbers by definition.
== Sources ==
* : Book $\text{VIII}$. Propositions
Category:Euclidean Number Theory
Numbers equal to Sum of Primes not Greater than its Prime Counting Function Value

== Theorem ==
Let $\map \pi n: \Z_{\ge 0} \to \Z_{\ge 0}$ denote the prime-counting function:
:$\map \pi n =$ the count of the primes less than $n$
Consider the equation:
:$\ds n = \sum_{p \mathop \le \map \pi n} p$
where $p \le \pi \left({n}\right)$ denotes the primes not greater than $\pi \left({n}\right)$.
Then $n$ is one of:
:$5, 17, 41, 77, 100$
== Proof ==
We have that:
== Sources ==
* 
* : $100$
Category:Prime Numbers
Category:Prime-Counting Function
Numbers equal to Sum of Primes not Greater than its Prime Counting Function Value/Mistake

== Source Work ==
:
:The Dictionary
::$100$
== Mistake ==
:The largest number for which the sum of the primes less than the number of primes less than or equal to the number is the number itself. In this case, $\map \pi n = 25$, and the sum of the primes from $2$ to $23 = 100$. The other numbers with this property are $5$, $17$, $41$ and $77$.
== Correction ==
The example given should have been presented as:
:$\map \pi {100} = 25$
But note that for $n = 5$ and $n = 17$, the sum of the primes less than $\map \pi 5 = 3$ and $\map \pi {17} = 7$ falls short of $5$ and $17$.
$3$ and $7$ themselves need to be included in that sum.
Hence the statement should be amended to:
:The largest number for which the sum of the primes less than or equal to the number of primes less than or equal to the number is the number itself.
== Sources ==
* : $100$
Category:Mistakes/Curious and Interesting Numbers
Numbers equal to Sum of Squares of two Parts

== Theorem ==
Integers that can be split into two parts whose squares add up to it include:
:$1233 = 12^2 + 33^2$
:$8833 = 88^2 + 33^2$
== Proof ==
== Sources ==
* 
* : $1233$
* : $1233$
Category:Sums of Squares
Category:Recreational Mathematics
Numbers for which Euler Phi Function equals Product of Digits

== Theorem ==
The sequence of positive integers $n$ for which $\map \phi n$ is equal to the product of the digits of $n$ begins:
:$1, 24, 26, 87, 168, 388, 594, 666, 1998, 2688, 5698, 5978, 6786, 7888, 68 \, 796$
It is known that this sequence is finite, but it is unknown whether a $16^{\text {th}}$ term exists.
== Proof ==
and so on.
Category:Euler Phi Function
Numbers for which Euler Phi Function of 2n + 1 is less than that of 2n

== Theorem ==
The sequence of positive integers for which:
:$\map \phi {2 n + 1} < \map \phi {2 n}$
begins:
:$157, 262, 367, 412, \ldots$
== Proof ==
== Sources ==
* : $157$
Category:Euler Phi Function
Numbers for which Sixth Power plus 1091 is Composite

== Theorem ==
The number $1091$ has the property that:
:$x^6 + 1091$
is composite for all integer values of $x$ from $1$ to $3905$.
== Proof ==
We check the result and show that it cannot be improved further by showing:
:$3906$ is the smallest $x$ such that $x^6 + 1091$ is prime.
Suppose $x^6 + 1091$ is prime.
Then:
:$x$ is a multiple of $42$
:$x$ ends in $0$, $4$ or $6$ in decimal notation
:$x \not \equiv \pm 1, \pm 3, \pm 4 \pmod {13}$
:$x \not \equiv \pm 4, \pm 6, \pm 9 \pmod {19}$
The proof is split into $6$ parts:
=== $x$ is a multiple of $2$ ===
Suppose not. Then $x$ is odd, and so is $x^6$.
Hence $x^6 + 1091$ is even, and thus composite.
Thus we must require $x$ to be even.
=== $x$ is a multiple of $3$ ===
Suppose not. Write $x = 3 k \pm 1$.
Hence:
showing that $x^6 + 1091$ is composite (divisible by $3$).
Thus we must require $x$ to be a multiple of $3$.
=== $x$ is a multiple of $7$ ===
Suppose not. Then $x \perp 7$.
Then:
showing that $x^6 + 1091$ is composite (divisible by $7$).
Thus we must require $x$ to be a multiple of $7$.
=== $x$ ends in $0$, $4$ or $6$ ===
From the above we require $x$ to be even.
Suppose $x$ ends in $2$ or $8$.
Then $x^6$ ends in $4$.
Thus $x^6 + 1091$ ends in $5$, which by Divisibility by 5 is divisible by $5$.
Hence we must require $x$ to end in $0$, $4$ or $6$.
=== $x \not \equiv \pm 1, \pm 3, \pm 4 \pmod {13}$ ===
Here is a table of $x^6 \pmod {13}$:
:$\begin{array}{|c|c|c|c|c|c|c|c|}
\hline x \bmod {13} & 0 & \pm 1 & \pm 2 & \pm 3 & \pm 4 & \pm 5 & \pm 6 \\
\hline x^6 \bmod {13} & 0 & 1 & -1 & 1 & 1 & -1 & -1 \\
\hline
\end{array}$
For $x \equiv \pm 1, \pm 3, \pm 4 \pmod {13}$:
showing that $x^6 + 1091$ is composite (divisible by $13$).
Thus we must require $x \not \equiv \pm 1, \pm 3, \pm 4 \pmod {13}$.
=== $x \not \equiv \pm 4, \pm 6, \pm 9 \pmod {19}$ ===
Here is a table of $x^6 \pmod {19}$:
:$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline x \bmod {19} & 0 & \pm 1 & \pm 2 & \pm 3 & \pm 4 & \pm 5 & \pm 6 & \pm 7 & \pm 8 & \pm 9 \\
\hline x^6 \bmod {19} & 0 & 1 & 7 & 7 & 11 & 6 & 11 & 1 & 1 & 11\\
\hline
\end{array}$
For $x \equiv \pm 4, \pm 6, \pm 9 \pmod {19}$:
showing that $x^6 + 1091$ is composite (divisible by $19$).
Thus we must require $x \not \equiv \pm 4, \pm 6, \pm 9 \pmod {19}$.
Below is a list of prime factorizations of $x^6 + 1091$, where $42 \divides x$ and does not end in $2$ or $8$.
If $x$ satisfies one of the above congruences, their remainder is listed instead.
The above results are obtained using the "Alpertron" Integer factorization calculator.
The whole operation took approximately $1$ second.
== Sources ==
* 
* 
* : $1091$
Category:Sixth Powers
Numbers forming Fraction in Lowest Terms are Coprime

== Theorem ==
== Proof ==
Let $A, B$ be (natural) numbers which are the least of those which have the same ratio with them.
We need to show that $A$ and $B$ are prime to one another.
:
 $A$ and $B$ are not coprime.
Then by definition there exists some (natural) number $C > 1$ which measures them both.
As many times as $C$ measures $A$, let that many units be in $D$.
As many times as $C$ measures $B$, let that many units be in $E$.
So by :
:$A = C \times D$
:$B = C \times E$
Thus by :
:$D : E = A : B$
So $D$ and $E$ are in the same ratio with $A$ and $B$.
But by Absolute Value of Integer is not less than Divisors: Corollary:
:$D < A$
:$E < B$
This contradicts our hypothesis that $A$ and $B$ are the least of those numbers that are in the same ratio with $A$ and $B$.
It follows that $A$ and $B$ must be prime to one another.
== Sources ==
* : Book $\text{VII}$. Propositions
Category:Coprime Integers
Numbers in Even-Even Amicable Pair are not Divisible by 3

== Theorem ==
Let $\tuple {m_1, m_2}$ be an amicable pair such that both $m_1$ and $m_2$ are even.
Then neither $m_1$ nor $m_2$ is divisible by $3$.
== Proof ==
An amicable pair must be formed from a smaller abundant number and a larger deficient number.
Suppose both $m_1, m_2$ are divisible by $3$.
Since both are even, they must also be divisible by $6$.
However $6$ is a perfect number.
By Multiple of Perfect Number is Abundant, neither can be deficient.
So $m_1, m_2$ cannot form an amicable pair.
Therefore at most one of them is divisible by $3$.
 suppose $m_1$ is divisible by $3$.
Write:
:$m_1 = 2^r 3^t a, m_2 = 2^s b$
where $a, b$ are not divisible by $2$ or $3$.
Then:
Since $m + n$ is not divisible by $3$, $s$ must be even.
Similarly, by Divisor Sum Function is Multiplicative, $t$ must also be even.
In particular, both $s, t$ are at least $2$.
Now write:
:$m_1 = 2^2 \cdot 3 k, m_2 = 2^2 \cdot l$
where $k, l$ are some integers.
By Multiple of Perfect Number is Abundant, $m_1$ is abundant number.
Therefore $m_2 > m_1$.
This leads to $l > 3 k \ge 3$.
By Abundancy Index of Product is greater than Abundancy Index of Proper Factors:
But:
which is a contradiction.
Therefore neither $m_1$ nor $m_2$ is divisible by $3$.
== Sources ==
* 
* : $220$
* : $220$
Category:Amicable Pairs