avewright/proofwiki-math huggingface dataset:
Theorem: Smallest Square Inscribed in Two Pythagorean Triangles

The smallest square with integer sides that can be inscribed within two different Pythagorean triangles so that one side of the square lies on the hypotenuse has side length $780$.
The two Pythagorean triangles in question have side lengths $\tuple {1443, 1924, 2405}$ and $\tuple {1145, 2748, 2977}$.

Proof:
By Inscribed Squares in Right-Angled Triangle/Side Lengths/Side Lies on Hypotenuse:
:For a Pythagorean triangle with side lengths $a, b, c$, the required inscribed square has side length given by:
::$\dfrac {abc}{ab + c^2}$
For primitive Pythagorean triples, $a, b, c$ are pairwise coprime, so the above fraction is in canonical form.
In other words, if the required side length is an integer, the triangle cannot be primitive, and this side length would be equal to some multiple of the product $abc$ of its primitive version.
Therefore in order to find the smallest such square, we would need to compare and find two sets of primitive Pythagorean triples such that their product would have a sufficiently small lowest common multiple.
The $\tuple {3, 4, 5}$ triple has a product of $60$.
The $\tuple {5, 12, 13}$ triple has a product of $780$.
These two products have a lowest common multiple of $780$.
Since $\sqrt [3] {780} < 10$, there is no need to search any further for triangles with smallest side length greater than $10$.
The ones remaining are:
:$7, 24, 25$
:$8, 15, 17$
:$9, 40, 41$
which all have products greater than $780$.
Therefore the solution must be generated with the two smallest triangles.
The inscribed square side length for the $\tuple {3, 4, 5}$ triangle is:
:$\dfrac {60} {3 \times 4 + 5^2} = \dfrac {60} {37}$
so it must be enlarged $481$ times to have a side length of $780$.
The inscribed square side length for the $\tuple {5, 12, 13}$ triangle is:
:$\dfrac {780} {5 \times 12 + 13^2} = \dfrac {780} {229}$
so it must be enlarged $229$ times to have a side length of $780$.
After enlargement, we get the triples $\tuple {1443, 1924, 2405}$ and $\tuple {1145, 2748, 2977}$, which is the result we have.
{{qed}}
Theorem: Sequence of Numbers Divisible by Sequence of Primes

The integers in this sequence:
:$788, 789, 790, 791, 792, 793$
are divisible by:
:$2, 3, 5, 7, 11, 13$
respectively.

Proof:
{{begin-eqn}}
{{eqn | l = 788
      | r = 2 \times 394
}}
{{eqn | l = 789
      | r = 3 \times 263
}}
{{eqn | l = 790
      | r = 5 \times 158
}}
{{eqn | l = 791
      | r = 7 \times 113
}}
{{eqn | l = 792
      | r = 11 \times 72
}}
{{eqn | l = 793
      | r = 13 \times 61
}}
{{end-eqn}}
{{qed}}
Theorem: Numbers whose Square is Palindromic with Even Number of Digits

The sequence of positive integers whose square is a palindromic number with an even number of digits begins:
:$836, 798 \, 644, 64 \, 030 \, 648, 83 \, 163 \, 115 \, 486, 6 \, 360 \, 832 \, 925 \, 898, \ldots$
{{OEIS|A016113}}

Proof:
{{begin-eqn}}
{{eqn | l = 836^2
      | r = 698 \, 896
      | c = $6$ digits
}}
{{eqn | l = 798 \, 644^2
      | r = 637 \, 832 \, 238 \, 736
      | c = $12$ digits
}}
{{eqn | l = 64 \, 030 \, 648^2
      | r = 4 \, 099 \, 923 \, 883 \, 299 \, 904
      | c = $16$ digits
}}
{{eqn | l = 83 \, 163 \, 115 \, 486^2
      | r = 6 \, 916 \, 103 \, 777 \, 337 \, 773 \, 016 \, 196
      | c = $22$ digits
}}
{{eqn | l = 6 \, 360 \, 832 \, 925 \, 898^2
      | r = 40 \, 460 \, 195 \, 511 \, 188 \, 111 \, 559 \, 106 \, 404
      | c = $26$ digits
}}
{{end-eqn}}
Theorem: Sum of Sequence of Factorials

The sequence $S = \sequence {s_n}$ defined as:
:$\ds s_n = \sum_{k \mathop = 1}^n k!$
begins:
:$1, 3, 9, 33, 153, 873, 5913, 46 \, 233, 409 \, 113, 4 \, 037 \, 913, \ldots$
{{OEIS|A007489}}

Proof:
{{begin-eqn}}
{{eqn | l = s_1
      | r = 1!
      | c = 
}}
{{eqn | r = 1
      | c = {{Defof|Factorial}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = s_2
      | r = s_1 + 2!
      | c = 
}}
{{eqn | r = 1 + 2
      | c = {{Defof|Factorial}}
}}
{{eqn | r = 3
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = s_3
      | r = s_2 + 3!
      | c = 
}}
{{eqn | r = 3 + 6
      | c = {{Defof|Factorial}}
}}
{{eqn | r = 9
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = s_4
      | r = s_3 + 4!
      | c = 
}}
{{eqn | r = 9 + 24
      | c = {{Defof|Factorial}}
}}
{{eqn | r = 33
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = s_5
      | r = s_4 + 5!
      | c = 
}}
{{eqn | r = 33 + 120
      | c = {{Defof|Factorial}}
}}
{{eqn | r = 153
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = s_6
      | r = s_5 + 6!
      | c = 
}}
{{eqn | r = 153 + 720
      | c = {{Defof|Factorial}}
}}
{{eqn | r = 873
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = s_7
      | r = s_6 + 7!
      | c = 
}}
{{eqn | r = 873 + 5040
      | c = {{Defof|Factorial}}
}}
{{eqn | r = 5913
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = s_8
      | r = s_7 + 8!
      | c = 
}}
{{eqn | r = 5913 + 40 \, 320
      | c = {{Defof|Factorial}}
}}
{{eqn | r = 46 \, 223
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = s_9
      | r = s_8 + 9!
      | c = 
}}
{{eqn | r = 46 \, 223 + 362 \, 880
      | c = {{Defof|Factorial}}
}}
{{eqn | r = 409 \, 113
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = s_{10}
      | r = s_9 + 10!
      | c = 
}}
{{eqn | r = 409 \, 113 + 3 \, 628 \, 800
      | c = {{Defof|Factorial}}
}}
{{eqn | r = 4 \, 037 \, 913
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:Factorials
Category:Sums of Sequences
i4bdq75wluxz4uzaozkqtbjihkqa5ld
Theorem: Primitive Semiperfect Numbers which are not Primitive Abundant

The sequence of primitive semiperfect numbers which are not also primitive abundant starts:
:$6, 28, 350, 490, 496, 770, 910, 1190, \ldots$
These are semiperfect numbers which are either:
: perfect
or:
: whose only abundant aliquot parts are weird.

Proof:
A primitive semiperfect number is a semiperfect number which has no aliquot parts which are themselves semiperfect.
Thus by definition a primitive semiperfect number is either perfect or abundant.
A primitive abundant number is an abundant number whose aliquot parts are all deficient.
Thus the perfect numbers:
:$6, 28, 496, \ldots$
are not primitive abundant.
However, by Divisor of Perfect Number is Deficient, the perfect numbers are all by definition primitive semiperfect.
Hence the presence of the perfect numbers in this sequence.
Next, consider the sequence of weird numbers:
:$70, 836, \ldots$
These are numbers which are abundant but not semiperfect.
Thus an abundant number whose aliquot parts are all deficient except for one or more weird numbers is a primitive semiperfect number which is not primitive abundant.
However, such a number is not primitive abundant because it has an aliquot part which is abundant, though weird.
So:
:while $70$ is primitive abundant it is not primitive semiperfect number
:while $350$, $490$, $710$, $910$, $1190$, and so on, are primitive semiperfect, they are not primitive abundant.
Hence the result.
{{qed}}
Category:Primitive Semiperfect Numbers
Category:Primitive Abundant Numbers
8aalkwm0o7fcvh38xcqvh31x3muet2s
Theorem: Set of 5 Triplets whose Sums and Products are Equal

The following set of $5$ triplets of integers have the property that:
:the sum of the integers in each triplet are equal
and:
:the product of the integers in each triplet are equal:
:$\tuple {6, 480, 495}$, $\tuple {11, 160, 810}$, $\tuple {12, 144, 825}$, $\tuple {20, 81, 880}$, $\tuple {33, 48, 900}$
The sum is $981$, and the product is $1 \, 425 \, 600$.
This is the only known such set of $5$ triplets of integers with this property.

Proof:
{{begin-eqn}}
{{eqn | l = 6 + 480 + 495
      | r = 981
}}
{{eqn | l = 11 + 160 + 810
      | r = 981
}}
{{eqn | l = 12 + 144 + 825
      | r = 981
}}
{{eqn | l = 20 + 81 + 880
      | r = 981
}}
{{eqn | l = 33 + 48 + 900
      | r = 981
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 6 \times 480 \times 495
      | r = \paren {2 \times 3} \times \paren {2^5 \times 3 \times 5} \times \paren {3^2 \times 5 \times 11}
}}
{{eqn | r = 2^6 \times 3^4 \times 5^2 \times 11
}}
{{eqn | r = 1 \, 425 \, 600
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 11 \times 160 \times 810
      | r = 11 \times \paren {2^5 \times 5} \times \paren {2 \times 3^4 \times 5}
}}
{{eqn | r = 2^6 \times 3^4 \times 5^2 \times 11
}}
{{eqn | r = 1 \, 425 \, 600
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 12 \times 144 \times 825
      | r = \paren {2^2 \times 3} \times \paren {2^4 \times 3^2} \times \paren {3 \times 5^2 \times 11}
}}
{{eqn | r = 2^6 \times 3^4 \times 5^2 \times 11
}}
{{eqn | r = 1 \, 425 \, 600
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 20 \times 81 \times 880
      | r = \paren {2^2 \times 5} \times 3^4 \times \paren {2^4 \times 5 \times 11}
}}
{{eqn | r = 2^6 \times 3^4 \times 5^2 \times 11
}}
{{eqn | r = 1 \, 425, 600
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 33 \times 48 \times 900
      | r = \paren {3 \times 11} \times \paren {2^4 \times 3} \times \paren {2^2 \times 3^2 \times 5^2}
}}
{{eqn | r = 2^6 \times 3^4 \times 5^2 \times 11
}}
{{eqn | r = 1 \, 425 \, 600
}}
{{end-eqn}}
{{qed}}
Theorem: Sum of Pandigital Triplet of 3-Digit Primes

The smallest integer which is the sum of a set of $3$ three-digit primes using all $9$ digits from $1$ to $9$ once each is $999$:
:$149 + 263 + 587 = 999$

Proof:
All three-digit primes end in $1, 3, 7, 9$.
Suppose $1$ is used as the units digit of a prime.
Since the digit $1$ cannot be used again, the sum of the primes is at least:
:$221 + 333 + 447 = 1001$
so $1$ cannot be used as a units digit .
The units digits of the primes are $3, 7, 9$.
To minimise the sum, the hundreds digits must be $1, 2, 4$.
This leaves $5, 6, 8$ be the tens digits.
The primes satisfying these conditions are:
:$157, 163, 167$
:$257, 263, 269, 283$
:$457, 463, 467, 487$
Only $269$ contain $9$, so we must choose it.
Only $157$ does not contain $6$, so must choose it next.
But then all primes beginning with $4$ have some digit coinciding with the above primes.
Hence the next minimal sum of the primes (if they exist) is:
:$10^2 \paren {1 + 2 + 5} + 10 \paren {4 + 6 + 8} + \paren {3 + 7 + 9} = 999$
and we have shown that these primes do exist.
{{qed}}
Theorem: Divisibility Test for 7, 11 and 13

Mark off the integer $N$ being tested into groups of $3$ digits.
Because of the standard way of presenting integers, this may already be done, for example:
:$N = 22 \, 846 \, 293 \, 462 \, 733 \, 356$
Number the groups of $3$ from the right: 
:$N = \underbrace{22}_6 \, \underbrace{846}_5 \, \underbrace{293}_4 \, \underbrace{462}_3 \, \underbrace{733}_2 \, \underbrace{356}_1$
Considering each group a $3$-digit integer, add the even numbered groups together, and subtract the odd numbered groups:
:$22 - 846 + 293 - 462 + 733 - 356 = -616$
where the sign is irrelevant.
If the result is divisible by $7$, $11$ or $13$, then so is $N$.
In this case:
:$616 = 2^3 \times 7 \times 11$
and so $N$ is divisible by $7$ and $11$ but not $13$.

Proof:
Let $N$ be expressed as:
:$N = \ds \sum_{k \mathop = 0}^n a_k 1000^k = a_0 + a_1 1000 + a_2 1000^2 + \cdots + a_n 1000^n$
where $n$ is the number of groups of $3$ digits.
We have that:
:$1000 \equiv -1 \pmod {1001}$
Hence from Congruence of Powers:
:$1000^r \equiv \paren {-1}^r \pmod {1001}$
Thus:
:$N \equiv a_0 + \paren {-1} a_1 + \paren {-1}^2 a_2 + \cdots + \paren {-1}^n a_n \pmod {1001}$
from the definition of Modulo Addition.
Then we note that:
:$1001 = 7 \times 11 \times 13$
and the result follows.
{{qed}}
Theorem: Solutions to Diophantine Equation x (x + 1) = y (y + 5) (y + 10) (y + 15)

The Diophantine equation
: $n = x \paren {x + 1} = y \paren {y + 5} \paren {y + 10} \paren {y + 15}$
has exactly $2$ solutions in $\N \setminus \set 0$:
{{begin-eqn}}
{{eqn | l = 1056
      | r = 32 \times 33 = 1 \times 6 \times 11 \times 16
}}
{{eqn | l = 43 \, 056
      | r = 207 \times 208 = 8 \times 13 \times 18 \times 23
}}
{{end-eqn}}

Proof:
First, we observe that:
{{begin-eqn}}
{{eqn | l = x \paren {x + 1}
      | r = \paren {x + \dfrac 1 2 }^2 - \paren {\dfrac 1 2}^2
}}
{{eqn | ll = \leadsto
      | l = x \paren {x + 1}
      | o = \lt
      | r = \paren {x + \dfrac 1 2 }^2
}}
{{eqn | ll = \leadsto
      | l = \sqrt {x \paren {x + 1} }
      | o = \lt
      | r = \paren {x + \dfrac 1 2 }
      | c = taking the square root of both sides
}}
{{end-eqn}}
We note here that the square root of the product $x \paren {x + 1}$ has a fractional part which is less than one half.
Therefore, the fractional part of the square root of the product of the four integers on the {{RHS}} must also be less than one half.
Next, we observe that:
{{begin-eqn}}
{{eqn | l = y \paren {y + 5} \paren {y + 10} \paren {y + 15}
      | r = \paren {y^2 + 15 y} \paren {y^2 + 15 y + 50}
}}
{{eqn | r = \paren {y^2 + 15 y} \paren {y^2 + 15 y + 25} + 25 \paren {y^2 + 15 y} 
      | c = 
}}
{{eqn | r = \paren {y^2 + 15 y} \paren {y^2 + 15 y + 25} + 25 \paren {y^2 + 15 y} + 25^2 - 25^2
      | c = adding $0$
}}
{{eqn | r = \paren {y^2 + 15 y} \paren {y^2 + 15 y + 25} + 25 \paren {y^2 + 15 y + 25} - 25^2
      | c = 
}}
{{eqn | r = \paren {y^2 + 15 y + 25}^2 - 25^2
      | c = 
}}
{{end-eqn}}
Let $a = \paren {y^2 + 15 y + 25}$
We know that $\sqrt {a^2 - 25^2 } \lt a$
We can narrow the search for solutions tremendously as follows:
We need $\sqrt {a^2 - 25^2 } \lt a - \dfrac 1 2$ so that the fractional part of $\sqrt {a^2 - 25^2 }$ '''might''' be less than one half.
Therefore:
{{begin-eqn}}
{{eqn | l = \sqrt {a^2 - 25^2 }
      | o = \lt
      | r = a - \dfrac 1 2
      | c = the fractional part of the square root of the product must be less than one half.
}}
{{eqn | ll = \leadsto
      | l = a^2 - 625
      | o = \lt
      | r = a^2 - a + \dfrac 1 4
      | c = squaring both sides
}}
{{eqn | ll = \leadsto
      | l = a
      | o = \lt
      | r = \dfrac {2501} 4
      | c = 
}}
{{end-eqn}}
Therefore:
{{begin-eqn}}
{{eqn | l = \paren {y^2 + 15 y + 25}
      | o = \lt
      | r = \dfrac {2501} 4
}}
{{eqn | ll = \leadsto
      | l = \paren {y^2 + 15 y + 25} - \dfrac {2501} 4
      | o = \lt
      | r = 0
      | c = 
}}
{{eqn | ll = \leadsto
      | l = \paren {y^2 + 15 y - \dfrac {2401} 4 }
      | o = \lt
      | r = 0
      | c = 
}}
{{eqn | ll = \leadsto
      | l = y
      | o = \lt
      | r = \dfrac {-15 \pm \sqrt {15^2 - 4 \times \paren {-\dfrac {2401} 4} } } 2
      | c = Quadratic Formula
}}
{{eqn | ll = \leadsto
      | l = y
      | o = \lt
      | r = \dfrac {\sqrt {2626 } - 15 } 2
      | c = 
}}
{{eqn | ll = \leadsto
      | l = y
      | o = \le
      | r = 18
      | c = 
}}
{{end-eqn}}
We now only need to review $18$ cases which are shown below.
:File:Diophantine.png
{{qed}}
Theorem: Numbers Reversed when Multiplying by 9

Numbers of the form $\sqbrk {10 (9) 89}_{10}$ are reversed when they are multiplied by $9$:
{{begin-eqn}}
{{eqn | l = 1089 \times 9
      | r = 9801
}}
{{eqn | l = 10 \, 989 \times 9
      | r = 98 \, 901
}}
{{eqn | l = 109 \, 989 \times 9
      | r = 989 \, 901
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \sqbrk {10 (9) 89}_{10}
      | r = 10 \times 10^{k + 2 } + 900 \sum_{i \mathop = 0}^{k - 1} 10^i + 89
      | c = {{Defof|Geometric Series}}
}}
{{end-eqn}}
Taking numbers of this form and multiplying by $9$ produces:
{{begin-eqn}}
{{eqn | l = 9 \times \paren {10 \times 10^{k + 2 } + 900 \sum_{i \mathop = 0}^{k - 1} 10^i + 89 }
      | r = 90 \times 10^{k + 2 } + 8100 \sum_{i \mathop = 0}^{k - 1} 10^i + 801
}}
{{end-eqn}}
The first part is composed of $k + 4$ digits. The first digit will be $9$ followed by $k + 3$ digits of $0$
{{begin-eqn}}
{{eqn | l = 90 \times 10^{k + 2 } 
      | r = 9 \times 10^{k + 3 } 
}}
{{end-eqn}}
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 $100$
{{begin-eqn}}
{{eqn | l = 8100 \sum_{i \mathop = 0}^{k - 1} 10^i
      | r = 899 \cdots 99100
}}
{{end-eqn}}
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 $01$:
{{begin-eqn}}
{{eqn | l = \sqbrk {10 (9) 89}_{10} \times 9
      | r = \sqbrk {98 (9) 01}_{10}
      | c =
}}
{{end-eqn}}
{{qed}}
Theorem: Reciprocal of 1089

:$\dfrac 1 {1089} = 0 \cdotp \dot 00091 \, 82736 \, 45546 \, 37281 \, 9 \dot 1$

Proof:
Performing the calculation using long division:
<pre>
     0.00091827364554637281910009...
    --------------------------------------------
1089)1.00000000000000000000000000...
       9801    5445    1089
       ----    ----    ----
        1990    5950    9910
        1089    5445    9801
        ----    ----    ----
         9010    5050    1090
         8712    4356    1089
         ----    ----    ----
          2980    6940      10000
          2178    6534       9801
          ----    ----       ----
           8020    4060      ....
           7623    3267
           ----    ----
            3970    7930
            3267    7623
            ----    ----
             7030    3070
             6534    2178
             ----    ----
              4960    8920
              4356    8712
              ----    ----
               6040    2080
               5445    1089        
</pre>{{qed}}
Theorem: Square which is Difference between Square and Square of Reversal

$33^2 = 65^2 - 56^2$
This is the only square of a $2$-digit number which has this property.

Proof:
{{begin-eqn}}
{{eqn | l = 33^2
      | r = 1089
      | c = 
}}
{{eqn | l = 65^2 - 56^2
      | r = 4225 - 3136
      | c = 
}}
{{eqn | r = 1089
      | c = 
}}
{{end-eqn}}
Let $\sqbrk {xy}$ be a $2$-digit number such that $x \ge y$ and $\sqbrk {xy}^2 - \sqbrk {yx}^2$ is a square of a $2$-digit number.
The case $x = y$ gives the solution $\sqbrk {xy}^2 - \sqbrk {yx}^2 = 0$, which is not a square of a $2$-digit number.
For $x \ne y$:
{{begin-eqn}}
{{eqn | l = \sqbrk {xy}^2 - \sqbrk {yx}^2
      | r = \paren {10 x + y}^2 - \paren {10 y + x}^2
}}
{{eqn | r = 100 x^2 + 20 x y + y^2 - 100 y^2 - 20 y x - x^2
}}
{{eqn | r = 99 \paren {x^2 - y^2}
}}
{{eqn | r = 3^2 \times 11 \paren {x - y} \paren {x + y}
}}
{{eqn | ll = \leadsto
      | l = \paren {x - y} \paren {x + y}
      | r = 11 n^2
      | c = for some integer $n$
}}
{{end-eqn}}
By Euclid's Lemma for Prime Divisors, one of $x - y, x + y$ must be divisible by $11$.
Hence from Absolute Value of Integer is not less than Divisors, either $x - y$ or $x + y$ must be greater than or equal to $11$.
Since $x - y < x < 9$, $x - y$ cannot be a multiple of $11$. 
From $1 = 1 + 0 \le x + y < 9 + 9 = 18$, we have that $x + y = 11$.
This implies that $x - y$ is a square number.
$x + y = 11$ gives $\tuple {x,y} = \tuple {6,5}, \tuple {7,4}, \tuple {8,3}, \tuple {9,2}$ as the possible solutions.
Among these solutions, only $\tuple {6,5}$ has a difference of a square number: $1$.
Therefore the only square of a $2$-digit number expressible as a difference between a square and the square of its reversal is:
:$65^2 - 56^2 = 33^2$.
{{qed}}
Theorem: Numbers for which Sixth Power plus 1091 is Composite

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:
Theorem: 1105 as Sum of Two Squares

$1105$ can be expressed as the sum of two squares in more ways than any smaller integer:
{{begin-eqn}}
{{eqn | l = 1105
      | m = 1089 + 16
      | mo= =
      | r = 33^2 + 4^2
      | c = 
}}
{{eqn | m = 1024 + 81
      | mo= =
      | r = 32^2 + 9^2
      | c = 
}}
{{eqn | m = 961 + 144
      | mo= =
      | r = 31^2 + 12^2
      | c = 
}}
{{eqn | m = 625 + 529
      | mo= =
      | r = 24^2 + 23^2
      | c = 
}}
{{end-eqn}}

Proof:
Here is the source code of a program in Python that finds all positive integers up to $1105$ that can be written as a sum of two squares in more ways than any smaller positive integer:
<syntaxhighlight lang="python">
import numpy as np
def two_sq_decomp_rich(n):
    bound = int(np.floor(np.sqrt(n)))
    count_of_two_sq_decomps = []
    for i in range(2*(bound + 1)*(bound + 1)):
        count_of_two_sq_decomps.append(0)
    for i in range(bound+1):        
        for j in range(i+1):
            count_of_two_sq_decomps[i*i+j*j] = count_of_two_sq_decomps[i*i+j*j] + 1
    max_sq_decomps = 0
    sq_decomp_rich_numbers = []
    for i in range(n+1):
        if count_of_two_sq_decomps[i] > max_sq_decomps:
            max_sq_decomps = count_of_two_sq_decomps[i]
            sq_decomp_rich_numbers.append(i)
    return sq_decomp_rich_numbers
print(two_sq_decomp_rich(1105))
</syntaxhighlight>
Output:
<syntaxhighlight lang="python">
  [0, 25, 325, 1105]
</syntaxhighlight>
[https://tio.run/##hZJLbsMgEED3nGKkbkwiRXGrbiK1p@guiixiQ0xkZjBgpT69C24@dRunXlgI3rz5gO1DTfgyDNpYcgGwM7YH4QEtY0/wUUtQ1DR00ngA1WEZNCEojZUH0TSJ30vnobMQCBBCLQKUAmEv4eR0CBIheqJQgO8MkIJwIvBtJ5z0oBEMuYiK3qdYBIE9eBPV0p3ljFVyjCp8W1SyJGMLp8s6Q75hEL/xt6cOK3iLxpChXamGyKWFb13c4JyPVBmpUJAqJjof47a7EVDkQKeynMCDzJ4X2bd4CTn/uT6nfuBcCWslVtma3xOPpmXONxfLVZfA4w3UCboePsi31Qu9PC6Ou9jMv0hs4TY6Iz6ns1iP29NpF5ebnhsVTgrVar6IHbz/yjlt8E8986ZJ3P2CL/egORtpJ0PncAZmzLr0gu68tjxfv3KeDMPwBQ Try it online!]
{{qed}}