# Find the minimum integer satisying certain condition

• Sep 12th 2010, 07:34 AM
Shanks
Find the minimum integer satisying certain condition
Find the minimum odd integer(positive) \$\displaystyle a(>5)\$such that:
(1)\$\displaystyle a=m_1^2+n_1^2\$ for some positive integers \$\displaystyle m_1,n_1\$.
(2)\$\displaystyle a^2=m_2^2+n_2^2\$ for some positive integers \$\displaystyle m_2,n_2\$.
(3)\$\displaystyle m_1-n_1=m_2-n_2\$.
• Sep 12th 2010, 09:47 AM
wonderboy1953
Quote:

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

• Sep 13th 2010, 05:39 AM
Shanks
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.
• Sep 13th 2010, 12:32 PM
Opalg
Quote:

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

From (2), \$\displaystyle (m_2,n_2,a)\$ is a Pythagorean triple. If this triple is primitive then there are integers s, t such that \$\displaystyle m_2 = s^2-t^2\$, \$\displaystyle n_2 = 2st\$ and \$\displaystyle a = s^2+t^2\$. So we can try taking \$\displaystyle m_1=s\$ and \$\displaystyle n_1 = t\$. In that case, (3) says that \$\displaystyle s-t = s^2-t^2 - 2st = (s-t)^2 - 2t^2\$. Therefore \$\displaystyle \tfrac12(s-t)(s-t-1) = t^2\$. This says that \$\displaystyle t^2\$ is a square triangular number. The smallest such number apart from 1 is \$\displaystyle 36 = 6^2 = \tfrac12(9)(8)\$. That suggests taking t = 6 and s – t = 9, so that s = 15.

That gives the solution \$\displaystyle a = 261 = 15^2+6^2\$, with \$\displaystyle 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.