Find all possible solutions of c and d

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Discussion Overview

The discussion revolves around finding all possible positive integer solutions for the variables c and d in the equation \(a^2 + b^2 + c^2 = d^2\), given specific values for a and b (70 and 61, respectively). The focus includes mathematical reasoning and exploration of potential solutions.

Discussion Character

  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • One participant presents the equation \(70^2 + 61^2 + c^2 = d^2\) and reformulates it to \(d^2 - c^2 = 8621\), suggesting that the factors of 8621 can be used to find pairs of (d+c) and (d-c).
  • Another participant claims to find a solution set of (d, c) = (4321, 4320) based on the factorization of 8621.
  • Further contributions reiterate the factorization approach, identifying pairs such as (8621, 1) and (233, 37) to derive additional solutions.
  • One participant acknowledges a calculation error and proposes two sets of solutions: (d, c) = (4311, 4310) and (135, 98), while also correcting earlier claims.

Areas of Agreement / Disagreement

Participants present multiple solutions for the values of c and d, indicating a lack of consensus on the correct pairs. Some solutions overlap, while others differ, suggesting ongoing exploration and verification of results.

Contextual Notes

There are indications of calculation errors and corrections throughout the discussion, highlighting the complexity of deriving integer solutions from the given equation.

albert391212
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$$if : a^2 + b^2 +c^2 = d^2 $$
where a,b c and d both are positive integers
if a=70 ,b=61
find all posible values of c and d
 
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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
 
82402F25-C005-4C6B-A029-D11AB6DD1936.jpeg
 
kaliprasad said:
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
 
(d+c)(d-c)=8621=(8621)(1)=(233)(37)
the solution is
(d,c)=(4321,4320),(135,98)#
 
kaliprasad said:
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 said:
(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)#
 

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