Find the minimum integer satisying certain condition

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September 12th 2010, 07:34 AM
Shanks

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) $m_1-n_1=m_2-n_2$ .
September 12th 2010, 09:47 AM
wonderboy1953

Quote:

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.
September 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.
September 13th 2010, 12:32 PM
Opalg

Quote:

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.