# 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) $m_1-n_1=m_2-n_2$.
I believe the answer to your first question is 13.

3. I think， you misread the problem, we are supposed to find the minimum number satisfying all the three condition. maybe that is my fault, but thanks anyway.

4. 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) $m_1-n_1=m_2-n_2$.
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.