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

Question Number 23105 by Rasheed.Sindhi last updated on 26/Oct/17

(∣ 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} \dots (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} \dots (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 .