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Diophantine Equations Sample

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The mathematician Diophantus of Alexandria around 250A. D. started some sort of research on some equations affecting more than one variables which would take merely whole number values. These equations are famously known as “DIOPHANTINE EQUATION” . named due to Diophantus. The simplest type of Diophantine equations that we shall see is the Linear Diophantine equations in two variables: ax+by=c. where a. b. degree Celsius are whole numbers and a. Bs are non both nothing. We besides have many sorts of Diophantine equations where our chief end is to happen out its solutions in the set of whole numbers.

Interestingly we can see some good theoretical treatment in Euclid’s “ELEMENTS” but no comment had been cited by Diophantus in his research works sing this type of equations.

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Whole Numbers

In figure theory. we are normally concerned with the belongingss of the whole numbers. or whole Numberss: Z = { . . . .3. 2. 1. 0. 1. 2. 3. . . . } . Let us get down with a really simple job that should be familiar to anyone who has studied simple algebra.

• Suppose that dolls sell for 7 dollars each. and toy train sets sell for 18 dollars. A storesells 25 sum dolls and train sets. and the entire sum received is 208 dollars. How many of each were sold?

The standard solution is straight-forward: Let x be the figure of dolls and Y be the figure of train sets. Then we have two equations and two terra incognitas: ten + y = 257x + 18y = 208The equations above can be solved in many ways. but possibly the easiest is to observe that the first one can be converted to: x = 25?y and so that value of x is substituted into the other equation and solved:

7 ( 25 ? Y ) + 18y = 208.i. e. 175 ? 7y + 18y = 208.i. e. ?7y + 18y = 208 ? 175.i. e. 11y = 33.i. e. Y = 3.Then if we substitute y = 3 into either of the original equations. we obtain ten = 22. and it is easy to look into that those values satisfy the conditions in the original job. Now let’s expression at a more interesting job:

Suppose that dolls sell for 7 dollars each. and toy train sets sell for 18 dollars. A shop sells merely dolls and train sets. and the entire sum received is 208 dollars. How many of each were theory?

This clip there is merely one equation: 7x+18y = 208. We likely learned in algebra category that you need as many equations as terra incognitas to work out jobs like this. so at first it seems hopeless. but there is one extra cardinal piece of information: the figure of dolls and the figure of train sets must be non-negative whole Numberss. With that in head. let’s see what we can make. disregarding for the minute the fact that we already have a solution. viz. : ten = 22 and y = 3.

Again. the all other solutions will be of the signifier given below: X=22+18cY=3-7c. where degree Celsius is any whole number

If we want to hold positive built-in solutions so by an easy calculation we observe that c= -1. 0. 1.When an equation of this kind is solvable by this method. there is no bound to the figure of stairss that need to be taken to obtain the solution. In the illustration above. we needed to present whole numbers a. B and c. but other equations might necessitate more or fewer of these intermediate values.

Linear Diophantine Equations

What we have merely solved is known as a Diophantine equation – an equation whose roots are required to be whole numbers. Probably the most celebrated Diophantine equation is the one stand foring Fermat’s last theorem. eventually proved 100s of old ages after it was proposed by AndrewWiles:

If n & 2. there are no non-trivial1 solutions in whole numbers to the equation: ten n + Y N = omega NThere are many. many signifiers of Diophantine equations. but equations of the kind that we merely solved are called “linear Diophantine equations” : all the coefficients of the variables are whole numbers. Let’s look a little more closely at the equation we merely solved: 7x + 18y = 208. If the merely requirementwere that the roots be whole numbers ( non needfully non-negative whole numbers ) . so our solution: ten = 22 + 18c and y = 3 ? 7c represent an infinte set of solutions. where every different whole number value of degree Celsius generates another solution. A more geometric position of the job is this: If we were to chart the equation 7x + 18y = 208. the solutions are topographic points where the graph passes through points that have integer co-ordinates. A part of that line is plotted. and the points where the graph has integer co-ordinates are indicated and labeled.

Notice that all the points with whole number co-ordinates are equally spaced along the line. In fact. if we begin at any point and add 18 to the x-coordinate and at the same clip subtract 7 from the y-coordinate. we arrive at another point on the graph with whole number co-ordinates. A speedy scrutiny of the original equation should do it obvious why this is the instance. The equation is:

7x + 18y = 208.If we add 18 to the x value. we increase the left side by 7 · 18. If we subtract 7 from the Y value. we decrease the left side by the same sum: 18 · 7. The net consequence is to go forth the left side unchanged.

Notice that this line has a negative incline and happens to cut through the first quarter-circle ( quadrant I ) and cross some points with whole number co-ordinates at that place. This may or may non be the instance for the graphs of other additive Diophantine equations. Lines with positive inclines can hold an infinite figure of solutions where both are positive. and there are equations where there are none. It’s easy to build such equations with whatever features you wish.

Does every equation of the signifier:ax + by = degree Celsius.where a. B and degree Celsius are whole numbers have a solution ( x. y ) . where ten and Y are besides whole numbers? The reply is no. For illustration. what if a and B are even and c is uneven? The left side must be even. and if the right side is uneven. there is no possibility of a solution with whole number values. Similarly. if a and B are both multiples of 3 and degree Celsiuss is non. the left side will be a multiple of 3 and the right side is non. so once more. there are no possible whole number solutions.

In fact. if the greatest common factor ( GCD ) of a and B does non split c. so there are no integer solutions. The astonishing thing. nevertheless. is that if the GCD of a and B besides divides c. so there are an infinite figure of integer solutions. and we will see why that is the instance subsequently on.

Note besides that another observation we made about our peculiar job will besides use to a general additive Diophantine equation ; viz. . that if ( x. Y ) is an integer solution to:ax + by = degree Celsiusso so will be ( ten + berkelium. y ? Alaska ) where K is any whole number. If we substitute ten + berkelium for ten and y ? Alaska for y. we obtain:a ( ten + berkelium ) + B ( y ? Alaska ) = degree Celsiusax + abk + by ? abk = degree Celsiusax + by = degree Celsius.so if ( x. Y ) is a solution. so so besides is ( ten + berkelium. y ? Alaska ) .

Euclid’s Algorithm and Diophantine Equations

Now let’s use the Euclidean algorithm on two of the Numberss from the original Diophantine equation we solved in Section 1: 7x + 18y = 208.18 = 7 · 2 + 47 = 4 · 1 + 34 = 3 · 1 + 13 = 1 · 3 + 0In this illustration. the concluding figure is 1. so the GCD of 18 and 7 is 1 ( in other words. 18 and 7 are comparatively premier ) . but the interesting thing to observe is that the Numberss in the GCD computation: 18. 7. 4. 3. 1 are the same Numberss that we got as denominators and as the coefficients of the variables in the numerators in the fractions when we were work outing the Diophantine equation 7x + 18y = 208. The lone eccentric Numberss were the invariables in the numerators. and that’s non surprising: we ne’er used the figure 208 when we were utilizing the Euclidean algorithm to happen the GCD of 18 and 7. If you check the arithmetic computations that are being done in each instance. it will be clear why the Numberss generated in both illustrations must be the same.

Suppose that the original Diophantine equation had had a 1 alternatively of the 208. To do certain you understand the technique we used to work out our Diophantine equation it would be a good exercising to work out the undermentioned equation by yourself before reading on: 7x + 18y = 1

The nice thing about the 1 in topographic point of the 208 is that it remains changeless throughout the computation. whereas the 208 was reduced as assorted of the denominators divided it equally. In this computation. all the other coefficients are the same as the Numberss generated in the straight-forward computation of the GCD of 7 and 18. To finish the solution. we need to back-substitute the B = 1  3c and after a few stairss we obtain: ten = 5 + 18c and y = 2  7c. where degree Celsius is an arbitrary whole number. ( Obviously this equation will hold no solutions where both x and Ys are positive. ) Therefore when you do a GCD computation of a and b. and that GCD turns out to be 1. you’ve done a batch of the work toward work outing the Diophantine equation ax + by = 1.

So if we can make the Euclidean algorithm. we can happen with about no attempt other than a small arithmetic the coefficients we need to work out a additive Diophantine equation of the signifier ax + by = 1. Of class we’d like to be able to work out equations where the 1 is replaced by an arbitrary figure c. but that is really non excessively hard. As as illustration. let’s find solutions for 7x + 18y = 208 presuming that we’ve solved 7x+ 18y = 1. The solutions for the latter equation are ten = 5+18c and y = 2 ?7c. where degree Celsius is an arbitrary whole number. An easy solution is merely to put hundred = 0 and obtain ten = 5 and y = 2 as a peculiar solution. But if we multiply ten and y by 208. so the left side will be increased by a factor of 208 so if we increase the right side by the same factor. we’ll have an ( x. Y ) brace that satisfies our original equation 7x + 18y = 208. Thus a solution is this: ten = ?5 · 208 = 1040 and y = 2 · 208 = 416. It’s easy to stop up these Numberss in to look into that they are valid.

But we besides noticed that adding any multiple of 18 to x while at the same clip adding that same multiple of ?7 to y will give the other solutions. so the general solution to our original job is: ten = ?1040+18k and y =416?7k. If thousand = 58. for illustration. this yields the solution x = 4 and y = 10.

We have seen that if we have any solution to one of these additive Diophantine equations. we can obtain all the others by adding changeless multiples of the opposite coefficients to the given solution. all we truly necessitate is one solution.

In the old illustrations. once we got to the point where we had b = 1?3c. we backsubstitutedand carefully kept path of the coefficient of degree Celsius in the computations. But since any solution will bring forth all the others. why non allow c = 0? Then we merely necessitate to track a individual figure.5. Puting It All Together:Let’s use the techniques above. but in their most simplified signifier. to work out another Diophantine equation. So expression at the job stated below:• In a pet store. rats cost 5 dollars. rainbow fishs cost 3 dollars and crickets cost 10 cents. One 100 animate beings are sold. and the entire grosss are 100 dollars. How many rats. rainbow fishs and crickets were sold?

If r. g and c represent the figure of rats. rainbow fishs and crickets. severally. we’ve got two equations ( but three terra incognitas ) :R + g + degree Celsius = 1005r + 3g + . 1c = 100To turn the job into a strictly whole number job. multiply the 2nd equation by 10: R + g + c = 10050r + 30g + degree Celsius = 1000If we subtract the first equation from the 2nd we obtain the familiar looking additive Diophantine equation in two variables:49r + 29g = 900.Fortunately. the GCD for 49 and 29 is 1 which divides 900 so there will be solutions ( although perchance non solutions where all the values are non-negative. ( This job is likely much easier to work out utilizing “guess and check” techniques: we know that the figure of crickets must be a multiple of 10. so you could merely seek 0. 10. 20. … . 100 of them. ) . Let’s find the GCD of 49 and 29. utilizing the Euclidian algorithm: 49 = 29 · 1 + 20

29 = 20 · 1 + 920 = 9 · 2 + 29 = 2 · 4 + 12 = 1 · 2 + 0

Therefore we get two whole numbers s & amp ; Ts such that 49s+29t=1.Then multiplying 900 to both the whole numbers s & amp ; Ts would give a solution to the original job.NEXT WE SHALL PUT SOME LIGHT ON THE QUADRATIC DIOPHANTINEEquations LIKE AS:1 ) ( ax+by ) ( cx+dy ) =e.2 ) axy=bx+cy+d.3 ) ax 2 +by 2 =cx+dy+e.4 ) ax 2 +by 2 =cz 2where a. b. c. d. vitamin E are all non-zero whole numbers.We can besides hold in our manus tonss of Diophantine equations which can non be included in our treatment.

Quadratic Equations

Next we will use the Unique Factorization Theorem to the solution of the diophantine equationx 2+ Y 2= z 2in whole numbers x. y. omega are whole numbers. Such three-base hits of solutions are called Pythagorean three-base hits. The most celebrated of these three-base hits is of class ( 3. 4. 5 ) . It is rather easy to give expressions for bring forthing such three-base hits: for illustration. take ten = 2mn. Y = m 2 ? n 2

and omega = M2 +n 2( particular instances were known to the Babylonians. the general instance occurs in Euclid ) . It is less straightforward to verify that there are no other solutions ( this was first done by the Arabs in the tenth century ) . Assume that ( x. y. omega ) is a Pythagorean three-base hit. If 500 divides two of these. it divides the 3rd. and so ( x/d. y/d. z/d ) is another Pythagorean three-base hit. We may therefore presume that x. Y and omega are pairwise coprime ; such three-base hits are called crude. In peculiar. precisely one of them is even. Claim 1. The even whole number must be one of ten or Y. In fact. if omega is even. so x and Ys are uneven. Writing ten = 2X + 1. Y = 2Y + 1 and z = 2Z. we find 4X 2

+ 4X + 4Y 2+ 4Y + 2 = 4Z 2: but the left manus side is non divisible by 4: contradiction.Exchanging ten and Y if necessary we may presume that ten is even. Now we transfer the linear job x 2 +y 2= z 2into a multiplicative 1 ( if we are tousage alone factorisation. we need merchandises. non amounts ) by composing ten 2 = omega 2 ?y 2=( omega ? Y ) ( z + Y ) .Claim 2. gcd ( z ? y. z + Y ) = 2. In fact. set 500 = gcd ( z ? y. z + Y ) . Then d divides omega ? Y and omega + y. hence their sum 2z and their difference 2y. Now gcd ( 2y. 2z ) = 2 gcd ( y. z ) = 2. so 500 | 2 ; on the other manus. 2 | vitamin D since omega ? Y and omega + Y are even since omega and Y are uneven. Thus vitamin D = 2 as claimed. Some Properties:

1 ) Let a. b be two co- premier whole numbers such that Bachelor of Arts is a square. Then a and B are squares. 2 ) Let a. b be two positive whole numbers with gcd ( a. B ) = vitamin D such that Bachelor of Arts is a square. Then a/d and b/d are squares.

Theorem: If ( x. y. omega ) is a crude Pythagorean three-base hit with x even. so there exist co-prime whole numbers m. Ns such that x = 2mn. Y = m 2 ? n 2andomega = M2+ n 2.Note that if Y is even. so the general solution is given by x = m 2 ? n 2.Y = 2mn and z = M2+ n 2. Furthermore. if we drop the status that the three-base hits be crude so the theorem continues to keep if we besides drop the status that the whole numbers m. n be comparatively premier.

Note that if Y is even. so the general solution is given by x = m2 ? n2. Y= 2mn and z = M2 + n2. Furthermore. if we drop the status that the three-base hits be crude so the theorem continues to keep if we besides drop the status that the whole numbers m. n be comparatively premier.

Lagrange’s Trick

The same technique we used for work outing x 2+ y 2 = z 2can be used to work outequations of the type x 2+ ay 2= z 2 merely compose the equation in the signifier ay 2=( omega ? x ) ( z + x ) and utilize alone factorisation.Equations like x 2 + y 2 = 2z 2at first seem intractable utilizing this attackbecause we can’t bring forth a difference of squares. Lagrange. nevertheless. saw that in this instance generation by 2 saves the twenty-four hours because ( 2z ) 2 = 2?22+ 2y 2=( ten + Y ) 2+ ( ten ? Y ) 2. hence ( 2z ? x ? Y ) ( 2z + x + Y ) = ( ten ? Y ) 2. and now the solution returns precisely as for Pythagorean three-base hits.Let us now show that we can make something similar for any equation of type AX 2+BY 2= CZ 2holding at least one solution. First. multiplying through byA shows that it is sufficient to see equations X 2+ aY 2 = bZ 2. Assumethat ( x. y. omega ) is a solution of this equation. Then( bzZ ) 2= bz 2 Ten 2+ abz 2 Y 2= ( x 2+ ay 2 ) X2+ ( ax 2+ a 2 y 2 ) Y 2= ( twenty + ayY ) 2 + a ( yX ? xY ) 2.Therefore a ( yX ? xY ) 2 = ( bzZ ) 2? ( twenty + ayY ) 2is a difference of squares. and wecan continue as for Pythagoren three-base hits. We have proved:Theorem: If the equation ax 2+ by 2= cz 2has a nontrivial solutionin whole numbers. so this equation can be factored over the whole numbers ( perchance after multiplying through by a suited whole number ) .

Problem: Find out a rectangle with same country and margin. if exists. Solution: Let a. B be the length and bredth of the rectangle. so we have by the undermentioned equation: 2 ( a+b ) =abThis is a Diophantine equation in a. B.We can work out it by utilizing Unique factorisation belongings as follows: ab-2a-2b=0.i. e. a ( b-2 ) -2 ( b-2 ) =4.i. e. ( a-2 ) ( b-2 ) =4connoting that either a-2=4. b-2=1 or a-2=1. b-2=4 or a-2=2. b-2=2 i. e. either a=6. b=3 or a=3. b=6 or a=4. b=4.Therefore we obtained two types of rectangles with the given belongings. Note that we can work out Diophantine equations utilizing geometry. alone factorisation belongings. inequalities and test methods

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Diophantine Equations Sample. (2017, Aug 17). Retrieved from https://graduateway.com/diophantine-equations-essay-sample-essay/

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