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8.2.1 math 2-a

 
 
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Re: 8.2.1 math 2-a
by David Reynoso - Saturday, April 25, 2020, 11:50 AM
 

Since $c = m^2 + n^2$, and $c < 30$, we have $m^2 + n^2 < 30$. Now, since $n^2 > 0$ this implies that $m^2 < 30$, so $m < \sqrt{30} < 6$. That is why the hint says it is enough to check values of $m$ and $n$ such that $0 < n < m < 6$. 

The idea then is to try all values of $m$ and $n$ with $0 < n < m  < 6$ and $\gcd(m.n) = 1$. Note not all of them will give you an answer, for example $n = 4$ and $m = 5$ would give $c = 4^2 + 5^2 = 41 >30$, but at least it narrows down the possible values of $m$ and $n$ that we need to check.