Find all possible solutions of c and d

[albert391212]

\(\displaystyle if : a^2 + b^2 +c^2 = d^2 \)
where a,b c and d both are positive integers
if a=70 ,b=61
find all posible values of c and d

 

[kaliprasad]

we have $70^2 + 61^2 + c^2 = d^2$
or $4900 +3721 + c^2 -= d^2$
or $d^2-c^2 = 8621$
or ($d+c)(d-c) = 8621$
to solve the same you can put 8621 as product of 2 numbers (one case 8621 * 1$ equiate one to d +c and another to d- c and solve for c and d. by choosing all pair of numbers fot erach pair one set of c d can be obtained

 

[skeeter]

 

[albert391212]

we have $70^2 + 61^2 + c^2 = d^2$
or $4900 +3721 + c^2 -= d^2$
or $d^2-c^2 = 8621$
or ($d+c)(d-c) = 8621$
to solve the same you can put 8621 as product of 2 numbers (one case 8621 * 1$ equiate one to d +c and another to d- c and solve for c and d. by choosing all pair of numbers fot erach pair one set of c d can be obtained

(d+c)(d-c)=8621
the only solution is
d=4321
c=4320

 

[albert391212]

(d+c)(d-c)=8621=(8621)(1)=(233)(37)
the solution is
(d,c)=(4321,4320),(135,98)#

 

[albert391212]

we have $70^2 + 61^2 + c^2 = d^2$
or $4900 +3721 + c^2 -= d^2$
or $d^2-c^2 = 8621$
or ($d+c)(d-c) = 8621$
to solve the same you can put 8621 as product of 2 numbers (one case 8621 * 1$ equiate one to d +c and another to d- c and solve for c and d. by choosing all pair of numbers fot erach pair one set of c d can be obtained

Thaks for your answer
Kaliprasad

Albert

 

[albert391212]

(d+c)(d-c)=8621=(8621)(1)=(233)(37)
the solution is
(d,c)=(4321,4320),(135,98)#

(d+c)(d-c)=8621=(8621)(1)=(233)(37)
the solution is
(d,c)=(4311,4310),(135,98)#
sorry again
I have a poor calculation
we have two sets of solution
d=135 or 4311
c=98 or 4310
(d,c)=(4311,4310),(135,98)#