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