Solve in positive integers for a² = 9555² + c²

In summary, the conversation discusses a math problem involving finding positive integer solutions for the equation a^2 = 9555^2 + c^2. One person offers a solution but is corrected by another person who points out that their solution only takes into account one aspect of the equation. The conversation ends with apologies and a discussion about the restrictions of the problem.
  • #1
mente oscura
168
0
Hello.

A simple question.Solve in all positive integers, for:

[tex]a^2=9555^2+c^2[/tex]

Please, you show the way of solving it.

Regards.
 
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  • #2
Re: Solve in positive integers for a^2=9555^2+c^2

$$a^2=9555^2+c^2$$
$$a^2-c^2=9555^2$$
$$(a-c)(a+c)=9555^2$$
If we let $f$ be a factor of $9555^2$, then we can let $a-c=f$ and $a+c=\frac{9555^2}{f}$, and $a=\frac{1}{2}(f+\frac{9555^2}{f})$. Now, since all the factors of $9555$ are odd, then $a$ will always be an integer. $9555$ has $24$ factors, hence there are $12$ possible pairs of $(a,c)$.
 
  • #3
Re: Solve in positive integers for a^2=9555^2+c^2

Hey mente oscura, I feel like I want to attack this easy challenge problem!:p:eek:
First notice that $a^2=9555^2+c^2$ can be rewritten in the following form:

$a^2-c^2=9555^2$

$(a+c)(a-c)=9555^2$

Since $9555=3\cdot5\cdot7\cdot7\cdot13$, if we are to arrange this number into the product of two square numbers, we see that we can have a total of $3+3+1+3+1=11$ ways to do so.

Those 11 ways can be formulated in the following manner:

1.

$(a+c)(a-c)=(3\cdot5)^2(7^2\cdot13)^2=15^2\cdot637^2=202997^2-202772^2$
$(a,c)=(202997, 202772)$
2.

$(a+c)(a-c)=(3\cdot7)^2(5\cdot7\cdot13)^2=21^2\cdot455^2=103733^2-103292^2$
$(a,c)=(103733, 103292)$
3.

$(a+c)(a-c)=(3\cdot13)^2(5\cdot7^2)^2=39^2\cdot245^2=30773^2-29252^2$
$(a,c)=(30773, 29252)$
4.

$(a+c)(a-c)=(5\cdot7)^2(3\cdot7\cdot13)^2=35^2\cdot273^2=37877^2-36652^2$
$(a,c)=(37877, 36652)$
5.

$(a+c)(a-c)=(5\cdot13)^2(3\cdot7^2)^2=65^2\cdot147^2=12917^2-8692^2$
$(a,c)=(12917, 8692)$
6.

$(a+c)(a-c)=(7\cdot13)^2(3\cdot5\cdot7)^2=91^2\cdot105^2=9653^2-1372^2$
$(a,c)=(9653, 1372)$
7.

$(a+c)(a-c)=(3\cdot5\cdot13)^2(7^2)^2=195^2\cdot49^2=20213^2-17812^2$
$(a,c)=(20213, 17812)$
8.

$(a+c)(a-c)=(3\cdot5\cdot7\cdot7)^2(13)^2=735^2\cdot13^2=270197^2-270028^2$
$(a,c)=(270197, 270028)$
9.

$(a+c)(a-c)=(3\cdot5\cdot7\cdot13)^2(7)^2=1365^2\cdot7^2=931637^2-931588^2$
$(a,c)=(931637, 931588)$
10.

$(a+c)(a-c)=(3\cdot7\cdot7\cdot13)^2(5)^2=1911^2\cdot5^2=1825973^2-1825948^2$
$(a,c)=(1825973, 1825948)$
11.

$(a+c)(a-c)=(5\cdot7\cdot7\cdot13)^2(3)^2=3185^2\cdot3^2=5072117^2-5072108^2$
$(a,c)=(5072117, 5072108)$
Edit: kaliprasad is so kind to inform me through PM that I missed to consider the number $1$ and hence missed out the case where

$(a+c)(a-c)=(3\cdot5\cdot7\cdot7\cdot13)^2(1)^2=9555^2\cdot1^2=45649013^2-45649012^2$ and the last solution set to the problem would be $(a,c)=(45649013, 45649012)$!

Thank you kaliprasad so much for your kind gesture!:eek:
 
Last edited:
  • #4
Re: Solve in positive integers for a^2=9555^2+c^2

$9555 = 3 * 5 * 7^2 * 13$
So $9555^2 = 3^2 * 5^2 * 7^4 * 13^2$
This has (2+1)(2+1)(4+1)(2+1) or 135 factors out of which one is 9555 and 67 are below 9555 and 67 are above 9555
(a + c) > 9555 and (a-c) < 9555 is a set of solution
So number of solutions 67
For a we need to take$ 3^x * 5^ y * 7 ^ z * 13^ m$ (x,y,z,m to be chosen based on limit for example x between 0 and 3 such that a > 9555) and for c it is $9555^2/a$
 
  • #5
Re: Solve in positive integers for a^2=9555^2+c^2

Hello.

Thank you very much to all for taking part.

Pythagorean numbers[tex]Let \ a, \ b, \ c, \ p, \ q \in{N} \ / \ a^2=b^2+c^2 \ and \ b=pq[/tex]

[tex]a=\dfrac{p^2+q^2}{2}[/tex]

[tex]b=pq[/tex]

[tex]c=\dfrac{p^2-q^2}{2}[/tex]
pqabc
1955545649013955545649012
33185507211795555072108
51911182597395551825948
713659316379555931588
137352701979555270028
156372029979555202772
214551037339555103292
3527337877955536652
3924530773955529252
4919520213955517812
651471291795558692
91105965395551372

Regards.
 
  • #6
Re: Solve in positive integers for a^2=9555^2+c^2

mente oscura said:
Hello.

Thank you very much to all for taking part.

Pythagorean numbers[tex]Let \ a, \ b, \ c, \ p, \ q \in{N} \ / \ a^2=b^2+c^2 \ and \ b=pq[/tex]

[tex]a=\dfrac{p^2+q^2}{2}[/tex]

[tex]b=pq[/tex]

[tex]c=\dfrac{p^2-q^2}{2}[/tex]
pqabc
1955545649013955545649012
33185507211795555072108
51911182597395551825948
713659316379555931588
137352701979555270028
156372029979555202772
214551037339555103292
3527337877955536652
3924530773955529252
4919520213955517812
651471291795558692
91105965395551372

Regards.

above ans is wrong as it takes care of b = pq only and not c = pq

my ans is right

the solution provided by computer program justifies the same

nt x = 9555* 9555;
#include <stdio.h>
main()
{
int a;
int b;
int i=0;
for ( a = 1; a < 9555; a ++)
{
b = x/a;
if( a * b == x) {
i++;
printf("the solution %i is %d %d\n", i, (a+b)/2, (b-a)/2);
}

}

}

the solution set is
the solution 1 is 45649013 45649012
the solution 2 is 15216339 15216336
the solution 3 is 9129805 9129800
the solution 4 is 6521291 6521284
the solution 5 is 5072117 5072108
the solution 6 is 3511469 3511456
the solution 7 is 3043275 3043260
the solution 8 is 2173773 2173752
the solution 9 is 1825973 1825948
the solution 10 is 1304275 1304240
the solution 11 is 1170507 1170468
the solution 12 is 1014445 1014400
the solution 13 is 931637 931588
the solution 14 is 724619 724556
the solution 15 is 702325 702260
the solution 16 is 608691 608616
the solution 17 is 501683 501592
the solution 18 is 434805 434700
the solution 19 is 390221 390104
the solution 20 is 310611 310464
the solution 21 is 270197 270028
the solution 22 is 260939 260764
the solution 23 is 234195 234000
the solution 24 is 202997 202772
the solution 25 is 186445 186200
the solution 26 is 167349 167076
the solution 27 is 145075 144760
the solution 28 is 140621 140296
the solution 29 is 133259 132916
the solution 30 is 103733 103292
the solution 31 is 100555 100100
the solution 32 is 90291 89784
the solution 33 is 87213 86688
the solution 34 is 78325 77740
the solution 35 is 71981 71344
the solution 36 is 62475 61740
the solution 37 is 56147 55328
the solution 38 is 54445 53600
the solution 39 is 47307 46332
the solution 40 is 44877 43848
the solution 41 is 39179 37996
the solution 42 is 37877 36652
the solution 43 is 34125 32760
the solution 44 is 30773 29252
the solution 45 is 29771 28196
the solution 46 is 27475 25760
the solution 47 is 24843 22932
the solution 48 is 21805 19600
the solution 49 is 21203 18928
the solution 50 is 20213 17812
the solution 51 is 19275 16740
the solution 52 is 17069 14144
the solution 53 is 16331 13244
the solution 54 is 15925 12740
the solution 55 is 14637 11088
the solution 56 is 14259 10584
the solution 57 is 13195 9100
the solution 58 is 12917 8692
the solution 59 is 12467 8008
the solution 60 is 11445 6300
the solution 61 is 10829 5096
the solution 62 is 10675 4760
the solution 63 is 10101 3276
the solution 64 is 9939 2736
the solution 65 is 9805 2200
the solution 66 is 9653 1372
the solution 67 is 9611 1036
 
  • #7
Re: Solve in positive integers for a^2=9555^2+c^2

kaliprasad said:
above ans is wrong as it takes care of b = pq only and not c = pq

my ans is right

the solution provided by computer program justifies the same

nt x = 9555* 9555;
#include <stdio.h>
main()
{
int a;
int b;
int i=0;
for ( a = 1; a < 9555; a ++)
{
b = x/a;
if( a * b == x) {
i++;
printf("the solution %i is %d %d\n", i, (a+b)/2, (b-a)/2);
}

}

}

the solution set is
the solution 1 is 45649013 45649012
the solution 2 is 15216339 15216336
the solution 3 is 9129805 9129800
the solution 4 is 6521291 6521284
the solution 5 is 5072117 5072108
the solution 6 is 3511469 3511456
the solution 7 is 3043275 3043260
the solution 8 is 2173773 2173752
the solution 9 is 1825973 1825948
the solution 10 is 1304275 1304240
the solution 11 is 1170507 1170468
the solution 12 is 1014445 1014400
the solution 13 is 931637 931588
the solution 14 is 724619 724556
the solution 15 is 702325 702260
the solution 16 is 608691 608616
the solution 17 is 501683 501592
the solution 18 is 434805 434700
the solution 19 is 390221 390104
the solution 20 is 310611 310464
the solution 21 is 270197 270028
the solution 22 is 260939 260764
the solution 23 is 234195 234000
the solution 24 is 202997 202772
the solution 25 is 186445 186200
the solution 26 is 167349 167076
the solution 27 is 145075 144760
the solution 28 is 140621 140296
the solution 29 is 133259 132916
the solution 30 is 103733 103292
the solution 31 is 100555 100100
the solution 32 is 90291 89784
the solution 33 is 87213 86688
the solution 34 is 78325 77740
the solution 35 is 71981 71344
the solution 36 is 62475 61740
the solution 37 is 56147 55328
the solution 38 is 54445 53600
the solution 39 is 47307 46332
the solution 40 is 44877 43848
the solution 41 is 39179 37996
the solution 42 is 37877 36652
the solution 43 is 34125 32760
the solution 44 is 30773 29252
the solution 45 is 29771 28196
the solution 46 is 27475 25760
the solution 47 is 24843 22932
the solution 48 is 21805 19600
the solution 49 is 21203 18928
the solution 50 is 20213 17812
the solution 51 is 19275 16740
the solution 52 is 17069 14144
the solution 53 is 16331 13244
the solution 54 is 15925 12740
the solution 55 is 14637 11088
the solution 56 is 14259 10584
the solution 57 is 13195 9100
the solution 58 is 12917 8692
the solution 59 is 12467 8008
the solution 60 is 11445 6300
the solution 61 is 10829 5096
the solution 62 is 10675 4760
the solution 63 is 10101 3276
the solution 64 is 9939 2736
the solution 65 is 9805 2200
the solution 66 is 9653 1372
the solution 67 is 9611 1036

It is not correct. You have had in account that ...?:

[tex]a=\dfrac{p^2+q^2}{2}[/tex]

Regards.
 
  • #8
Re: Solve in positive integers for a^2=9555^2+c^2

It is correct. $(a, c) = (9805, 2200)$ cannot be seen in either your or anemone's solution.
 
  • #9
Re: Solve in positive integers for a^2=9555^2+c^2

mente oscura said:
It is not correct. You have had in account that ...?:

[tex]a=\dfrac{p^2+q^2}{2}[/tex]

Regards.

I would like to see with evidence why it is not correct
 
  • #10
Re: Solve in positive integers for a^2=9555^2+c^2

kaliprasad said:
above ans is wrong as it takes care of b = pq only and not c = pq

my ans is right

the solution provided by computer program justifies the same

nt x = 9555* 9555;
#include <stdio.h>
main()
{
int a;
int b;
int i=0;
for ( a = 1; a < 9555; a ++)
{
b = x/a;
if( a * b == x) {
i++;
printf("the solution %i is %d %d\n", i, (a+b)/2, (b-a)/2);
}

}

}

the solution set is
the solution 1 is 45649013 45649012
the solution 2 is 15216339 15216336
the solution 3 is 9129805 9129800
the solution 4 is 6521291 6521284
the solution 5 is 5072117 5072108
the solution 6 is 3511469 3511456
the solution 7 is 3043275 3043260
the solution 8 is 2173773 2173752
the solution 9 is 1825973 1825948
the solution 10 is 1304275 1304240
the solution 11 is 1170507 1170468
the solution 12 is 1014445 1014400
the solution 13 is 931637 931588
the solution 14 is 724619 724556
the solution 15 is 702325 702260
the solution 16 is 608691 608616
the solution 17 is 501683 501592
the solution 18 is 434805 434700
the solution 19 is 390221 390104
the solution 20 is 310611 310464
the solution 21 is 270197 270028
the solution 22 is 260939 260764
the solution 23 is 234195 234000
the solution 24 is 202997 202772
the solution 25 is 186445 186200
the solution 26 is 167349 167076
the solution 27 is 145075 144760
the solution 28 is 140621 140296
the solution 29 is 133259 132916
the solution 30 is 103733 103292
the solution 31 is 100555 100100
the solution 32 is 90291 89784
the solution 33 is 87213 86688
the solution 34 is 78325 77740
the solution 35 is 71981 71344
the solution 36 is 62475 61740
the solution 37 is 56147 55328
the solution 38 is 54445 53600
the solution 39 is 47307 46332
the solution 40 is 44877 43848
the solution 41 is 39179 37996
the solution 42 is 37877 36652
the solution 43 is 34125 32760
the solution 44 is 30773 29252
the solution 45 is 29771 28196
the solution 46 is 27475 25760
the solution 47 is 24843 22932
the solution 48 is 21805 19600
the solution 49 is 21203 18928
the solution 50 is 20213 17812
the solution 51 is 19275 16740
the solution 52 is 17069 14144
the solution 53 is 16331 13244
the solution 54 is 15925 12740
the solution 55 is 14637 11088
the solution 56 is 14259 10584
the solution 57 is 13195 9100
the solution 58 is 12917 8692
the solution 59 is 12467 8008
the solution 60 is 11445 6300
the solution 61 is 10829 5096
the solution 62 is 10675 4760
the solution 63 is 10101 3276
the solution 64 is 9939 2736
the solution 65 is 9805 2200
the solution 66 is 9653 1372
the solution 67 is 9611 1036

Sorry.

You two have the reason, there are absent Pythagorean compound numbers.

A restriction is absent in the terms of reference of the question, that asks for " primitive solutions ". And nonetheless, they would exceed 4 of the solutions that are compound.(4ª, 7ª, 8ª, 12ª)

Thousand excuses.:eek:

Regards.
 

FAQ: Solve in positive integers for a² = 9555² + c²

What is the value of a?

The value of a cannot be determined without knowing the value of c.

How do I solve for a in this equation?

To solve for a, you will need to use the Pythagorean theorem and solve for c first. Once you have the value of c, you can plug it into the equation a² = 9555² + c² and solve for a.

Can a have a negative value?

No, a cannot have a negative value since it is squared in the equation. The solution for a will always be a positive integer.

Is there more than one solution for a?

Yes, there are multiple solutions for a. This is because there can be multiple values of c that satisfy the equation a² = 9555² + c². However, all solutions for a will be positive integers.

Can this equation be solved without using the Pythagorean theorem?

No, the Pythagorean theorem is necessary to solve this equation. It is a fundamental mathematical principle that states in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

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