How many pairs of positive integer a, b

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Discussion Overview

The discussion revolves around finding pairs of positive integers \( a \) and \( b \) such that \( a^2 + b^2 = 121 \). Participants explore related mathematical expressions and the conditions under which they yield perfect squares.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant asks how many pairs of positive integers \( a \) and \( b \) satisfy \( a^2 + b^2 = 121 \).
  • Another participant inquires if there are positive integers for which the expression \( (11-b)(11+b) \) results in perfect squares.
  • A participant explains their reasoning for using the expression \( (11-b)(11+b) \) by referencing the subtraction of \( b^2 \) from both sides of the equation.
  • There is a suggestion that there may only be one solution, though this is not confirmed.
  • Participants discuss the implications of one of the integers being 11, noting that the other would need to be 0, which does not meet the criteria of being a positive integer.
  • One participant expresses uncertainty about how to find the values of \( a \) and \( b \) and acknowledges the definition of zero.
  • Another participant concludes that they cannot find a positive integer \( b \) such that \( a \) is a perfect square, suggesting there may be no solutions for \( a \) and \( b \).

Areas of Agreement / Disagreement

Participants do not reach a consensus on the number of pairs of positive integers \( a \) and \( b \) that satisfy the equation. There are competing views on the existence of such pairs and the conditions under which they can be found.

Contextual Notes

Participants express uncertainty regarding the existence of positive integers that satisfy the conditions of the problem, and there are unresolved mathematical steps in determining the pairs.

omicron
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How many pairs of positive integer a, b are such that [tex]a^2 + b^2 = 121?[/tex]
 
Last edited:
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Are there any positive integers for which

[tex](11-b)(11+b)[/tex]

are perfect squares?
 
How did u come out with (11-b)(11+b)?
 
I subtracted [itex]b^2[/itex] from both sides and factored the right side and the left side is a perfect square as you indicated.
 
So I guess there is only one?
 
What would that be?
 
erm... 11?
 
Last edited:
If one is 11 then the other has to be 0 but you specified positive integers.
 
Ok so how do i get it?
Btw i know that 0 isn't a positive integer, so what is it called then?
 
  • #10
if u cannot find b such that (11-b)(11+b) is a perfect square, what do u call such a situation?

0 is simply called zero.
However some literature do put zero in both positive and negative and indicate them separately as +0 and -0.+0 is positive zero and -0 in negative zero.

-- AI
 
  • #11
I don't know what u call such a situation.
Anyway back to the question. Do u know how many pairs of positive integer a, b are such that [tex]a^2+b^2=121[/tex]
 
  • #12
a = (11-b)(11+b)
i cannot find positive integer b such that a is a perfect square
therefore there is no solution for a and b.

-- AI
 
  • #13
Ok. Thanks!
 

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