nyuuzyou/wikis huggingface dataset:
Square on Second Bimedial Straight Line applied to Rational Straight Line

== Theorem ==
== Proof ==
:
Let $AB$ be a second bimedial straight line divided into its medials at $C$.
Let $AC > CB$.
Let $DE$ be a rational straight line.
Let $DEFG$ equal to $AB^2$ be applied to $DE$ producing $DG$ as its breadth.
It is to be demonstrated that $DG$ is a third binomial straight line.
From :
:$AB^2 = AC^2 + CB^2 + 2 \cdot AC \cdot CB$
Let the rectangle $DH$ be applied to $DE$ such that $DH = AC^2$.
Let the rectangle $KL$ be applied to $DE$ such that $KL = BC^2$.
Then the rectangle $MF$ is equal to $2 \cdot AC \cdot CB$.
Let $MG$ be bisected at $N$.
Let $NO$ be drawn parallel to $ML$ (or $GF$, which is the same thing).
Therefore each of the rectangles $MO$ and $NF$ equals $AC \cdot CB$.
We have that $AB$ is a second bimedial which has been divided into its medials at $C$.
Therefore, by definition, $AC$ and $CB$ are medial straight lines which are commensurable in square only such that $AC \cdot CB$ is a medial rectangle.
Thus, by definition, $AC^2$ and $CB^2$ are also medial.
From:
:
and:
:
it follows that:
:$DL$ is medial.
We have that $DL$ has been applied to the rational straight line $DE$.
Therefore from :
:$MD$ is rational and incommensurable in length with $DE$.
For the same reason:
:$MG$ is also rational straight line and incommensurable in length with $ML$, which equals $DE$.
Therefore each of $DM$ and $MG$ is rational straight line and incommensurable in length with $DE$.
We have that $AC$ is incommensurable in length with $CB$.
Also:
:$AC : CB = AC^2 : AC \cdot CB$
Therefore by :
:$AC^2$ is incommensurable with $AC \cdot CB$.
Hence by:
:
and:
:
it follows that:
:$AC^2 + CB^2$ is incommensurable with $2 \cdot AC \cdot CB$.
That is:
:$DL$ is incommensurable with $MF$.
So from:
:
and:
:
it follows that:
:$DM$ is incommensurable in length with $MG$.
But $DM$ and $MG$ are rational straight lines which are incommensurable in length with each other.
Therefore by definition $DG$ is binomial.
It remains to be proved that $DG$ is a third binomial straight line.
From :
:$AC^2 + CB^2 > 2 \cdot AC \cdot CB$
Therefore $DL > MF$.
From :
:$DM > MG$
Since:
:$AC^2$ is commensurable with $CB^2$
it follows that:
:$DH$ is commensurable with $KL$.
So from:
:
and:
:
it follows that:
:$DK$ is commensurable in length with $KM$.
Also $DK \cdot KM = MN^2$.
Thus from :
:$DM^2$ is greater than $MG^2$ by the square on a straight line commensurable in length with $DM$.
Also:
:Neither $DM$ nor $MG$ is commensurable in length with $DE$.
Therefore $DG$ is a third binomial straight line.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Square on Side of Equilateral Triangle inscribed in Circle is Triple Square on Radius of Circle

== Theorem ==
== Proof ==
:
Let $ABC$ be a circle.
Let the equilateral triangle $ABC$ be inscribed within the circle $ABC$.
It is to be demonstrated that the square on one side of $\triangle ABC$ is $3$ times the square on the radius of the circle $ABC$.
Let $D$ be the center of the circle $ABC$.
Let $AD$ be joined and produced to $E$.
Let $BE$ be joined.
We have that $\triangle ABC$ is equilateral.
Therefore the arc $BEC$ is a third part of the circumference of the circle $ABC$.
Therefore the arc $BE$ is a sixth part of the circumference of the circle $ABC$.
Therefore $BE$ is the side of a regular hexagon inscribed within the circle $ABC$.
Therefore from :
:$BE = DE$
We have that:
:$AE = 2 \cdot DE$
So:
:$AE^2 = 4 \cdot ED^2 = 4 \cdot BE^2$
But from :
:
and:
:
it follows that:
:$AE^2 = AB^2 + BE^2 = 4 \cdot BE^2$
Therefore:
:$AB^2 = 3 \cdot BE^2$
But:
:$BE = DE$
Therefore:
:$AB^2 = 3 \cdot DE^2$
Hence the result.
== Sources ==
* : Book $\text{XIII}$. Propositions
Category:Equilateral Triangles
Category:Circles
Square on Side of Rational plus Medial Area applied to Rational Straight Line

== Theorem ==
== Proof ==
:
Let $AB$ be the side of a rational plus a medial area divided at $C$.
Let $AC > CB$.
Let $DE$ be a rational straight line.
Let $DEFG$ equal to $AB^2$ be applied to $DE$ producing $DG$ as its breadth.
It is to be demonstrated that $DG$ is a fifth binomial straight line.
From :
:$AB^2 = AC^2 + CB^2 + 2 \cdot AC \cdot CB$
Let the rectangle $DH$ be applied to $DE$ such that $DH = AC^2$.
Let the rectangle $KL$ be applied to $DE$ such that $KL = BC^2$.
Then the rectangle $MF$ is equal to $2 \cdot AC \cdot CB$.
Let $MG$ be bisected at $N$.
Let $NO$ be drawn parallel to $ML$ (or $GF$, which is the same thing).
Therefore each of the rectangles $MO$ and $NF$ equals $AC \cdot CB$.
We have that $AB$ is the side of a rational plus a medial area which has been divided at $C$.
Therefore, by definition, $AC$ and $CB$ are straight lines which are incommensurable in square such that:
:$AC^2 + CB^2$ is medial
:$AC \cdot CB$ is a rational rectangle.
Since $AC^2 + CB^2$ is medial, $DL$ is medial.
Therefore from :
:$DM$ is rational and incommensurable in length with $DE$.
We have that $2 \cdot AC \cdot CB$, that is $MF$ is rational.
Therefore from :
:$MG$ is rational and commensurable in length with $DE$.
Therefore from :
$DM$ is incommensurable in length with $MG$
That is, $DM$ and $MG$ are rational straight lines which are commensurable in square only.
Therefore by definition $DG$ is binomial.
It remains to be proved that $DG$ is a fifth binomial straight line.
From :
:$AC^2 + CB^2 > 2 \cdot AC \cdot CB$
Therefore $DL > MF$.
From :
:$DM > MG$
Also $DK \cdot KM = MN^2$.
Because:
:$AC^2$ is incommensurable with $CB^2$
it follows that:
:$DH$ is incommensurable with $KL$.
So from:
:
and:
:
it follows that:
:$DK$ is incommensurable in length with $KM$.
So from :
:$DM^2$ is greater than $MG^2$ by the square on a straight line incommensurable in length with $DM$.
Also:
:$DM$ and $MG$ are rational straight lines which are commensurable in square only with each other
and:
:$DM$ is commensurable in length with the rational straight line $DE$.
Therefore $DG$ is a fifth binomial straight line.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Square on Side of Regular Pentagon inscribed in Circle equals Squares on Sides of Hexagon and Decagon inscribed in same Circle

== Theorem ==
== Proof ==
:
Let $ABCDE$ be a circle.
Let the regular pentagon $ABCDE$ be inscribed within the circle $ABCDE$.
It is to be demonstrated that the square on the side of the pentagon $ABCDE$ equals the square on the side of a regular hexagon plus the square on the side of a regular decagon which have both been inscribed within the circle $ABCDE$.
Let the center of the circle $ABCDE$ be $F$.
Let $AF$ be joined and produced to $G$.
Let $FB$ be joined.
Let $FH$ be drawn from $F$ perpendicular to $AB$ and produced to $K$.
Let $AK$ and $KB$ be joined.
Let $FL$ be drawn from $F$ perpendicular to $AK$ and produced to $M$.
Let $KN$ be joined.
We have that the arc $ABCG$ equals the arc $AEDG$.
Within those arcs, the arc $ABC$ equals the arc $AED$.
Therefore the remainders are equal: the arc $CG$ equals the arc $GD$.
But $CD$ is the side of a regular pentagon.
Therefore $CG$ (when joined) is the side of a regular decagon.
We have that:
:$FA = FB$
and:
:$FH \perp AB$
Therefore from:
:
and:
::
it follows that:
:$\angle AFK = \angle KFB$
Hence from :
:the arc $AK$ equals the arc $KB$.
Therefore:
:the arc $AB$ equals double the arc $BK$.
Therefore $AK$ is the side of a regular decagon.
For the same reason:
:the arc $AK$ equals double the arc $KM$.
We have that:
:the arc $AB$ equals double the arc $BK$
while:
:the arc $CD$ equals the arc $KB$.
Therefore:
:the arc $CD$ equals double the arc $BK$.
But:
:the arc $AK$ equals double the arc $KM$.
Therefore:
:the arc $BK$ equals double the arc $KM$
and therefore:
:the arc $CG$ equals double the arc $KM$.
But we have that:
:the arc $CB$ equals the arc $BA$
therefore
:the arc $CB$ equals double the arc $BK$.
Therefore:
:the arc $GB$ equals double the arc $BM$.
So from :
:$\angle GFB = 2 \cdot \angle BFM$
But we also have:
:$\angle FAB = \angle ABF$
and so:
:$\angle GFB = 2 \cdot \angle FAB$
Therefore:
:$\angle BFN = \angle FAB$
But $\angle ABF$ is common between $\triangle ABF$ and $\triangle BFN$.
Therefore from :
:$\angle AFB = \angle BNF$
Therefore from :
:$AB : BF = FB : BN$
Therefore from :
:$AB \cdot BN = BF^2$
We have that:
:$AL = LK$
while $LN$ is common and perpenducular to $AK$.
Therefore from :
:$KN = AN$
Therefore:
:$\angle LKN = \angle LAN$
But:
:$\angle LAN = \angle KBN$
Therefore:
:$\angle LAN = \angle KBN$
We have that:
:$\angle A$ is common to $\triangle AKB$ and $\triangle AKN$.
Therefore from :
:$\angle AKB = \angle KNA$
Therefore:
:$\triangle AKB$ is equiangular with $\triangle KNA$.
Therefore from :
:$BA : AK = KA : AN$
Therefore from :
:$BA \cdot AN = AK^2$
But we have that:
:$AB \cdot BN = BF^2$
Therefore:
:$AB \cdot BN + BA \cdot AN = BF^2 + AK^2$
But from :
:$AB \cdot BN + BA \cdot AN = BA^2$
That is:
:$BA^2 = BF^2 + AK^2$
But:
:$BA$ is the side of the pentagon $ABCDE$
:$AK$ is the side of a regular decagon inscribed within the circle $ABCDE$.
and from :
:$BF$ is the side of a regular hexagon inscribed within the circle $ABCDE$.
Hence the result.
== Sources ==
* : Book $\text{XIII}$. Propositions
Category:Pentagons
Category:Hexagons
Category:Decagons
Square on Side of Sum of two Medial Area applied to Rational Straight Line

== Theorem ==
== Proof ==
:
Let $AB$ be the side of the sum of two medial areas divided at $C$.
Let $AC > CB$.
Let $DE$ be a rational straight line.
Let $DEFG$ equal to $AB^2$ be applied to $DE$ producing $DG$ as its breadth.
It is to be demonstrated that $DG$ is a sixth binomial straight line.
From :
:$AB^2 = AC^2 + CB^2 + 2 \cdot AC \cdot CB$
Let the rectangle $DH$ be applied to $DE$ such that $DH = AC^2$.
Let the rectangle $KL$ be applied to $DE$ such that $KL = BC^2$.
Then the rectangle $MF$ is equal to $2 \cdot AC \cdot CB$.
Let $MG$ be bisected at $N$.
Let $NO$ be drawn parallel to $ML$ (or $GF$, which is the same thing).
Therefore each of the rectangles $MO$ and $NF$ equals $AC \cdot CB$.
We have that $AB$ is the side of the sum of two medial areas divided at $C$.
Therefore, by definition, $AC$ and $CB$ are straight lines which are incommensurable in square such that:
:$AC^2 + CB^2$ is medial
:$AC \cdot CB$ is a medial rectangle
:$AC^2 + CB^2$ is incommensurable with $AC \cdot CB$.
Since $AC^2 + CB^2$ is medial, $DL$ is medial.
Therefore from :
:$DM$ is rational and incommensurable in length with $DE$.
We have that:
:$2 \cdot AC \cdot CB$, which equals $MF$, is medial.
We also have that $MF$ is applied to the rational straight line $ML$.
Therefore from :
:$MG$ is rational and incommensurable in length with $DE$.
Therefore $DM$ and $MG$ are rational straight lines which are incommensurable in length with $DE$.
We have that $AC^2 + CB^2$ is incommensurable with $2 \cdot AC \cdot CB$.
Thus $DL$ is incommensurable with $MF$.
So from:
:
and:
:
it follows that:
:$DM$ is incommensurable in length with $MG$.
That is, $DM$ and $MG$ are rational straight lines which are commensurable in square only.
Therefore by definition $DG$ is binomial.
It remains to be proved that $DG$ is a sixth binomial straight line.
From :
:$AC^2 + CB^2 > 2 \cdot AC \cdot CB$
Therefore $DL > MF$.
From :
:$DM > MG$
Also $DK \cdot KM = MN^2$.
Because:
:$AC^2$ is incommensurable with $CB^2$
it follows that:
:$DH$ is incommensurable with $KL$.
So from:
:
and:
:
it follows that:
:$DK$ is incommensurable in length with $KM$.
So from :
:$DM^2$ is greater than $MG^2$ by the square on a straight line incommensurable in length with $DM$.
Also, neither $DM$ nor $MG$ is commensurable in length with the rational straight line $DE$.
Therefore $DG$ is a sixth binomial straight line.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Square on Straight Line which produces Medial Whole with Rational Area applied to Rational Straight Line

== Theorem ==
== Proof ==
:
Let $AB$ be a straight line which produces with a rational area a medial whole.
Let $CD$ be a rational straight line.
Let the rectangle $CE$ be applied to $CD$ equal to $AB^2$ producing $CF$ as breadth.
It is to be demonstrated that $CF$ is a fifth apotome.
Let $BG$ be the annex to $AB$.
Therefore, by definition, $AG$ and $GB$ are straight lines which are incommensurable in square which make $AG^2 + GB^2$ medial but $2 \cdot AG \cdot GB$ rational.
Let the rectangle $CH$ be applied to $CD$ equal to the square on $AG$, producing $CK$ as breadth.
Let the rectangle $KL$ be applied to $CD$ equal to the square on $BG$, producing $KM$ as breadth.
Then the whole $CL$ is equal to the squares on $AG$ and $GB$.
But $AG^2 + GB^2$ medial.
Therefore $CL$ is medial.
We have that $CL$ is applied to the rational straight line $CD$ producing $CM$ as breadth.
Therefore by :
:$CM$ is rational and incommensurable in length with $CD$.
We have that:
:$CL = AG^2 + GB^2$
and:
:$AB^2 = CE$
Therefore by :
:$2 \cdot AG \cdot GB = FL$
Let $FM$ be bisected at the point $N$.
Let $NO$ be drawn through $N$ parallel to $CD$.
Therefore each of the rectangles $FO$ and $LN$ is equal to the rectangle contained by $AB$ and $GB$.
We have that $2 \cdot AG \cdot GB$ is rational.
Therefore $FL$ is rational.
Also $FL$ is applied to the rational straight line $FE$, producing $FM$ as breadth.
Therefore from :
:$FM$ is rational and commensurable in length with $CD$.
We have that $CL$ is medial and $FL$ is rational.
Therefore $CL$ is incommensurable with $FL$.
But from :
:$CL : FL = CM : FM$
Therefore by :
:$CM$ is incommensurable in length with $FM$.
But both $CM$ and $FM$ are rational.
Therefore $CM$ and $FM$ are rational straight lines which are commensurable in square only.
Therefore, by definition, $CF$ is an apotome.
It remains to be shown that $CF$ is a fifth apotome.
It can be proved similarly that:
:$CK \cdot KM = NM^2 = \dfrac {FM^2} 4$
We have that $AG^2$ is incommensurable with $GB^2$.
We also have:
:$CH = AG^2$
and:
:$KL = BG^2$
Therefore $CH$ is incommensurable with $KL$.
But from :
:$CH : KL = CK : KM$
Therefore by :
:$CK$ is incommensurable with $KM$.
We have that:
:$CM$ and $MF$ are unequal straight lines
and:
:the rectangle $CK \cdot KM$ has been applied to $CM$ equal to $\dfrac {FM^2} 4$ and deficient by a square figure
while:
:$CK$ is incommensurable with $KM$.
Therefore from :
:$CM^2$ is greater than $MF^2$ by the square on a straight line which is incommensurable in length with $CM$.
Also the annex $FM$ is commensurable in length with the rational straight line $CD$.
Therefore, by definition, $CF$ is a fifth apotome.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Square which is Difference between Square and Square of Reversal

== Theorem ==
$33^2 = 65^2 - 56^2$
This is the only square of a $2$-digit number which has this property.
== Proof ==
Let $\sqbrk {xy}$ be a $2$-digit number such that $x \le y$ 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 $x \ne y$:
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:
:$65^2 - 56^2 = 33^2$.
== Sources ==
* : $1089$
* : $1089$
Category:Square Numbers
Category:Reversals
Square whose Divisor Sum is Cubic

== Theorem ==
The number $1 \, 857 \, 437 \, 604$ is a square number whose divisor sum is a cube.
== Proof ==
== Sources ==
* 
* : $1,857,437,604$
* : $1,857,437,604$
Category:Divisor Sum Function
Category:Square Numbers
Category:Cube Numbers
Category:1,857,437,604
Category:Integers whose Divisor Sum is Cube
Square whose Perimeter equals its Area

== Theorem ==
The $4 \times 4$ square is the only square whose area in square units equals its perimeter in units.
The area and perimeter of this square are $16$.
== Proof ==
Let $S$ be a square whose area equals its perimeter.
Let $A$ be the area of $S$.
Let $P$ be the perimeter of $S$.
Let $b$ be the length of one side of $S$.
From Area of Square:
:$A = b^2$
From Perimeter of Rectangle:
:$P = 2 b + 2 b = 4 b$
Setting $A = P$
:$b^2 = 4 b$
and so:
:$b = 4$
and so:
:$A = 16 = P$
== Historical Note ==
== Sources ==
* : $16$
* : $16$
Category:Squares
Squares Ending in 5 Occurrences of 2-Digit Pattern

== Theorem ==
Let $n$ be a square number whose decimal representation ends in the pattern $\mathtt {xyxyxyxyxy}$.
Then $\mathtt {xy}$ is one of:
:$21, 29, 61, 69, 84$
The smallest examples of such numbers are:
== Proof ==
=== Case $1$: $\mathtt {xy}$ is odd ===
Consider the last $3$ digits of $n$: $\mathtt {yxy}$.
By Odd Square Modulo 8:
:$n \equiv \mathtt {yxy} \equiv 1 \pmod 8$
By Square Modulo 5:
:$n \equiv \mathtt {yxy} \equiv 0, 1, 4 \pmod 5$
We have:
Since $y$ is odd, we have $\mathtt {yxy} \equiv 1 \mathtt {xy} \pmod {200}$.
Hence $\mathtt {xy}$ can only be $05, 21, 29, 45, 61, 69, 85$.
Suppose $n$ ends in $5$ in decimal representation.
By Divisibility by 5, $5 \divides n$.
Since $5$ is a prime, $5 \divides \sqrt n$.
By Odd Square Modulo 8 and Chinese Remainder Theorem:
:$n \equiv 25 \pmod {200}$
for which none of $\mathtt {xy} = 05, 45, 85$ satisfy.
This shows that $\mathtt {xy}$ can only be $21, 29, 61, 69$.
=== Case $2$: $\mathtt {xy}$ is even ===
Consider the last $2$ digits of $n$: $\mathtt {xy}$.
We omit the trivial case $\mathtt {xy} = 00$.
This case occurs whenever $10 \divides \sqrt n$.
Since $\sqrt n$ is even:
:$n \equiv \mathtt {xy} \equiv 0 \pmod 4$
Since $5 \nmid \sqrt n$, by Square Modulo 5:
:$n \equiv \mathtt {xy} \equiv 1, 4 \pmod 5$
We have:
In the list above:
:$4, 24, 36, 44, 56, 76, 84$ are divisible by $2^2$ but not $2^4$
:$16, 96$ are divisible by $2^4$ but not $2^6$
:$64$ is divisible by $2^6$
Write $n = 10^{10} k + 101010101 \times \mathtt {xy}$.
We require $\dfrac n {2^{2 d}}$ to be a square number as well, where $d$ is the largest integer where $2^{2d} \divides \mathtt {xy}$ (shown above).
We have:
and:
:$\mathtt {xy} = 04, 16, 24, 36, 44, 56, 64, 76, 84, 96$
corresponds to:
Thus the only valid $\mathtt {xy}$ is $84$.
We have considered all possibilities, and the only possible $\mathtt {xy}$ are:
:$21, 29, 61, 69, 84$
and by the examples shown, all these numbers do indeed produce squares.
== Historical Note ==
== Sources ==
* : $21$
* : $21$
Category:Number Theory
Category:Recreational Mathematics
Squares Ending in 5 Occurrences of 2-Digit Pattern/Historical Note

== Historical Note on Squares Ending in 5 Occurrences of 2-Digit Pattern ==
According to in his $1997$ book , this result is attributed to , in volume $6$ of the .
No online corroboration of this can be found online, as there appears to be no consolidated archive of this magazine.
The author of this page would give his right arm to have a copy of the entire series resting on a bookshelf in his study.
Category:Historical Notes
Squares Ending in 5 Occurrences of 2-Digit Pattern/Mistake

== Source Work ==
:
:The Dictionary
::$21$
== Mistake ==
:If a square ends in the pattern $xyxyxyxyxy$, then $xy$ is either $21$, $61$ or $84$.
== Correction ==
Also there are $29$ and $69$.
In , this has been corrected.
== Sources ==
* : $21$
Category:Mistakes/Curious and Interesting Numbers
Squares Ending in Repeated Digits

== Theorem ==
A square number $n^2$ can end in a repeated digit either:
:$(1): \quad n^2$ is a multiple of $100$, in which case $n$ is a multiple of $10$
:$(2): \quad n^2$ ends in $44$ and $n$ ends in $12, 38, 62$ or $88$.
== Proof ==
Let $n \in \Z_{>0}$ end in $a b$.
By the Basis Representation Theorem, $n$ can be expressed as:
:$n = 100 k + 10 a + b$
for some $k \in \Z_{>0}$ and for $0 \le a < 10, 0 \le b < 10$.
Then:
Thus the nature of the last $2$ digits of $n^2$ are not dependent upon $k$.
So we can ignore all digits of $n$ except the last $2$.
Note that if $b = 0$ we have that $b^2 = 0$ and $20 a b = 0$.
So the last $2$ digits of the square of a multiple of $10$ are both $0$, and $n^2$ is a multiple of $100$.
Let $a = 5 + c$ where $0 \le c < 5$.
We have that:
So the square of a number ending in $c b$, where $0 \le c < 5$, ends in the same $2$ digits as the square a number ending in $a b$ where $a = c + 5$.
It remains to list the integers from $1$ to $49$, generating their squares and investigating their last $2$ digits:
It is seen that the only $n^2$ ending in a repeated digit end in $44$.
The corresponding $n$ are seen to be $12$ and $38$.
Adding $5$ to the $10$s digit of each gives us $62$ and $88$ as further such:
The result follows.
== Sources ==
* : $144$
* : $144$
Category:Square Numbers
Squares Ending in n Occurrences of m-Digit Pattern/Example

== Example of Squares Ending in n Occurrences of m-Digit Pattern ==
The $n$th term in the sequence:
:$611, 734 \, 611, 494 \, 734 \, 611, 63 \, 494 \, 734 \, 611, \dots$
have squares ending in $n$ occurrences of $321$:
== Proof ==
First we note that $611^2 = 373 \, 321 \equiv 1 \, 321 \pmod {2 \, 000}$.
Hence the condition in the theorem is satisfied.
Now we construct the sequence using the Induction Step as an algorithm:
For $n = 2$, we find $b < 1 \, 000$ such that:
:$611 b \equiv \dfrac {321 - 1} 2 - \dfrac {373 \, 321 - 1 \, 321} {2 \, 000} \equiv -26 \pmod {10^3}$
we have $b = 234$.
However:
:$234 \, 611^2 = 55 \, 042 \, 321 \, 321 \equiv 0 \, 321 \, 321 \pmod {2 \times 10^6}$
So we need to take $b' = 234 + 500 = 734$.
We see that:
:$734 \, 611^2 = 539 \, 653 \, 321 \, 321 \equiv 1 \, 321 \, 321 \pmod {2 \times 10^6}$
For $n = 3$, we find $b < 1 \, 000$ such that:
:$734 \, 611 b \equiv \dfrac {321 - 1} 2 - \dfrac {539 \, 653 \, 321 \, 321 - 1, 321 \, 321} {2 \, 000 \, 000} \equiv -269 \, 666 \pmod {10^3}$
That is:
:$611 b \equiv 334 \pmod {10^3}$
we have $b = 994$.
However:
:$994 \, 734 \, 611^2 = 989 \, 496 \, 946 \, 321 \, 321 \, 321 \equiv 0 \, 321 \, 321 \, 321 \pmod {2 \times 10^9}$
So we need to take $b' = 994 - 500 = 494$.
We see that:
:$494 \, 734 \, 611^2 = 244 \, 762 \, 335 \, 321 \, 321 \, 321 \equiv 1 \, 321 \, 321 \, 321 \pmod {2 \times 10^9}$
For $n = 4$, we find $b < 1 \, 000$ such that:
:$494 \, 734 \, 611 b \equiv \dfrac {321 - 1} 2 - \dfrac {244 \, 762 \, 335 \, 321 \, 321 \, 321 - 1, 321 \, 321 \, 321} {2 \, 000 \, 000 \, 000} \equiv -122 \, 381 \, 007 \pmod {10^3}$
That is:
:$611 b \equiv -7 \pmod {10^3}$
we have $b = 563$.
However:
:$563 \, 494 \, 734 \, 611^2 = 317 \, 526 \, 315 \, 934 \, 321 \, 321 \, 321 \, 321 \equiv 0 \, 321 \, 321 \, 321 \, 321 \pmod {2 \times 10^{12}}$
So we need to take $b' = 563 - 500 = 63$.
We see that:
:$63 \, 494 \, 734 \, 611^2 = 4 \, 031 \, 581 \, 323 \, 321 \, 321 \, 321 \, 321 \equiv 1 \, 321 \, 321 \, 321 \, 321 \pmod {2 \times 10^9}$
The sequence can be continued.
Category:Number Theory
Category:Recreational Mathematics
Category:Squares Ending in n Occurrences of m-Digit Pattern
Squares equal to Sum of 2 Cubes

== Sequence ==
The sequence of integers whose square can be expressed as the sum of $2$ coprime cubes begins:
:$3, 228, 671, 1261, 6371, 9765, 35 \, 113, 35 \, 928, 40 \, 380, 41 \, 643, 66 \, 599, \ldots$
== Proof ==
== Sources ==
* : $51,984$
Category:Square Numbers
Category:Cube Numbers
Squares of Diagonals of Parallelogram

== Theorem ==
Let $ABCD$ be a parallelogram.
:
Then:
:$AC^2 + BD^2 = AB^2 + BC^2 + CD^2 + DA^2$
== Proof ==
But we have that $\angle C$ and $\angle B$ are supplementary:
:$\angle C = \angle B = 180 \degrees$
Thus from Cosine of Supplementary Angle:
:$\cos \angle B + \cos \angle C = 0$
The result follows.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $127$
Category:Parallelograms