Question-31594

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 .

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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, \dots, \pm 46}

For no value of m 2179 - m^{2} is perfect

square .