# Thread: Find the minimum integer satisying certain condition

1. ## Find the minimum integer satisying certain condition

Find the minimum odd integer(positive) $a(>5)$such that:
(1) $a=m_1^2+n_1^2$ for some positive integers $m_1,n_1$.
(2) $a^2=m_2^2+n_2^2$ for some positive integers $m_2,n_2$.
(3) $m_1-n_1=m_2-n_2$.

2. Originally Posted by Shanks
Find the minimum odd integer(positive) $a(>5)$such that:
(1) $a=m_1^2+n_1^2$ for some positive integers $m_1,n_1$.
(2) $a^2=m_2^2+n_2^2$ for some positive integers $m_2,n_2$.
(3) $m_1-n_1=m_2-n_2$.
Find the minimum odd integer(positive) $a(>5)$such that:
(1) $a=m_1^2+n_1^2$ for some positive integers $m_1,n_1$.
(2) $a^2=m_2^2+n_2^2$ for some positive integers $m_2,n_2$.
(3) $m_1-n_1=m_2-n_2$.
From (2), $(m_2,n_2,a)$ is a Pythagorean triple. If this triple is primitive then there are integers s, t such that $m_2 = s^2-t^2$, $n_2 = 2st$ and $a = s^2+t^2$. So we can try taking $m_1=s$ and $n_1 = t$. In that case, (3) says that $s-t = s^2-t^2 - 2st = (s-t)^2 - 2t^2$. Therefore $\tfrac12(s-t)(s-t-1) = t^2$. This says that $t^2$ is a square triangular number. The smallest such number apart from 1 is $36 = 6^2 = \tfrac12(9)(8)$. That suggests taking t = 6 and s – t = 9, so that s = 15.
That gives the solution $a = 261 = 15^2+6^2$, with $a^2 = 261^2 = 189^2 + 180^2$. This certainly satisfies all the conditions (1), (2), (3), but I do not know whether it is the smallest solution apart from a = 5.