Knowing $2017 = 9^2 + 44^2$. find $m,n$ which satisfy $2017^2 = m^2 + n^2$

Question

Given that $$2017 = 9^2 + 44^2,$$ use this relation to find at least one group of positive integers $m$ and $n$ that satisfy $$2017^2 = m^2 + n^2.$$

https://proofwiki.org/wiki/Brahmagupta-Fibonacci_Identity – lab bhattacharjee Jan 20 '20 at 08:47 — lab bhattacharjee, Jan 20 '20 at 08:47

score 3 · Accepted Answer · answered Jan 20 '20 at 08:34

3

Hint : $$(a^2+b^2)(c^2+d^2)=(ad+bc)^2+(ac-bd)^2$$

answered Jan 20 '20 at 08:34

Peter

84,454

That is Lagrange's identity… – Bernard Jan 20 '20 at 09:31

score 2 · Answer 2 · answered Jan 20 '20 at 09:09

2

To find one group of pairs $m,n$ just recall Pythagorean triplet $(a^{2}+b^{2})^{2}=(a^{2}-b^{2})^{2} + (2ab)^{2}$ Moreover note that $2017^{2}=792^{2}+1855^{2}$. That's it cheers!!

answered Jan 20 '20 at 09:09

Knowing $2017 = 9^2 + 44^2$. find $m,n$ which satisfy $2017^2 = m^2 + n^2$

2 Answers2