(∣m^2 −n^2 ∣,2mn,m^2 +n^2 ) is pythagorean triplet for all m,n∈N. This can be proved easily.Is the vice versa of the statement is also true? I-E If for a,b,c∈N ,a^2 +b^2 =c^2 then there exist m,n∈N such that m^2 +n^2 =c and {a,b}={∣m^2 −n^2 ∣,2mn}


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Question Number 23105 by Rasheed.Sindhi last updated on 26/Oct/17
$(∣m^2 −n^2 ∣,2mn,m^2 +n^2 ) is pythagorean triplet for all m,n∈N. This can be proved easily.Is the vice versa of the statement is also true? I-E If for a,b,c∈N ,a^2 +b^2 =c^2 then there exist m,n∈N such that m^2 +n^2 =c and {a,b}={∣m^2 −n^2 ∣,2mn}$
$(∣ m^{2} - n^{2} ∣, 2 mn, m^{2} + n^{2}) is pythagorean$ $triplet for all m, n \in N . This can be$ $proved easily . Is the vice versa of$ $the statement is also true ?$ $I - E$ $If for a, b, c \in N, a^{2} + b^{2} = c^{2} then there$ $exist m, n \in N such that m^{2} + n^{2} = c and$ ${a, b} = {∣ m^{2} - n^{2} ∣, 2 mn}$
Commented by prakash jain last updated on 27/Oct/17

${(m^{2} - n^{2})}^{2} + 4 m^{2} n^{2} = {(m^{2} + n^{2})}^{2}$ $a = m^{2} - n^{2}$ $b = 4 m^{2} n^{2}$ $a = m^{2} - n^{2}$ $c = m^{2} + n^{2}$ $m = \sqrt{\frac{c + a}{2}}, n = \sqrt{\frac{c - a}{2}}$ $c a s e (i)$ $Let us say a a n d c a r e o d d and b is even .$ $c = 2 k + 1$ $a = 2 l + 1$ $c^{2} - a^{2} = 4 (k^{2} - l^{2}) + 4 (k - l)$ $= 4 (k - l) (k + l - 1)$ $c + a = 2 (k + l + 1)$ $c - a = 2 (k - l)$ $b = 2 \sqrt{(k - l) (k + l + 1)}$ $a = (m^{2} - n^{2}), b = 4 m^{2} n^{2} if$ $to prove : (k - l) and (k + l + 1) are both whole$ $squares .$ $it is easy to see if (k - l) is whole$ $square then (k + l + 1) has to be$ $a whole square since b is an integer .$ $assume k - l = h^{2} p where p is not$ $whole square then (k + l + 1) = j^{2} p$ $k = \frac{j^{2} p + h^{2} p - 1}{2}, l = \frac{j^{2} p - h^{2} p - 1}{2} . . . (A)$ $We can easily see below if p$ $e x i s t s i n (A) t h e n p i s a c o m m o n$ $f a c t o r of a, b, c$ $k = \frac{j^{2} p + h^{2} p - 1}{2}, l = \frac{j^{2} p - h^{2} p - 1}{2} . . . (A)$ $a = 2 l + 1 = (j^{2} - h^{2}) p$ $c = 2 k + 1 = (j^{2} + h^{2}) p$ $So p exists as in (A) then p is a common$ $factor of (a, b, c)$ $Summary$ $Every pythagorean triple (a, b, c)$ $can be written in the form$ $(m^{2} - n^{2}, 2 m n, m^{2} - n^{2}) if (a, b, c)$ $have no common factor$ $where m = \sqrt{\frac{c - a}{2}}, n = \sqrt{\frac{c + a}{2}}$
Commented by prakash jain last updated on 26/Oct/17

$c a s e (i i)$ $c e v e n a n d a a n d b o d d$ $a = 2 k + 1$ $b = 2 i + 1$ $c = 2 l$ ${(2 k + 1)}^{2} + {(2 i + 1)}^{2} = 4 l^{2}$ $4 k^{2} + 4 k + 1 + 4 i^{2} + 4 k + 1 = 4 l^{2}$ $k^{2} + k + i^{2} + i + \frac{1}{2} = l^{2}$ $n o t p o s s i b l e$
Commented by Rasheed.Sindhi last updated on 27/Oct/17

$Sir thanks for a nice answer!$ $BTW Sir, why do you seem very rarely$ $in the forum - activities nowadays ?$
Commented by prakash jain last updated on 27/Oct/17

$Past few months i was very busy$ $in my office . So did not get any$ $time at all .$
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