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Question Number 31594 by Nayon.Sm last updated on 11/Mar/18

	Answered by Rasheed.Sindhi last updated on 11/Mar/18
	$a=0 ∣ b=0 ; No other integral solution. −.−.−.−.−.−.− If a^2 +b^2 =c^2 then there exist integers m & n such that c=m^2 +n^2 ,{a,b}={m^2 -n^2 ,2mn} So we have m^2 +n^2 =c=2179 n^2 =2179−m^2 I-e 2179−m^2 is perfect square of an integer. m^2 ≤2179⇒m≤46 m∈{0,±1,±2,±3,...,±46} For no value of m 2179−m^2 is perfect square.$
	$a = 0 ∣ b = 0; No other integral solution .$ $- . - . - . - . - . - . -$ $If a^{2} + b^{2} = c^{2} then there exist integers$ $m & n such that c = m^{2} + n^{2}, {a, b} = {m^{2} - n^{2}, 2 m n}$ $So we have m^{2} + n^{2} = c = 2179$ $n^{2} = 2179 - m^{2}$ $I - e 2179 - m^{2} is perfect square of an$ $integer .$ $m^{2} ⩽ 2179 \Rightarrow m ⩽ 46$ $m \in {0, \pm 1, \pm 2, \pm 3, . . ., \pm 46}$ $For no value of m 2179 - m^{2} is perfect$ $square .$
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