Find p and q such that p^2 +q^2 =101^2 . Where p, q∈Z different from zero.


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Question Number 113689 by Ar Brandon last updated on 14/Sep/20

$Find p and q such that$ $p^{2} + q^{2} = 101^{2} . Where p, q \in Z$ $different from zero .$
Commented by Rasheed.Sindhi last updated on 14/Sep/20

$p = 20, q = 99 o r p = 99, q = 20$
	Answered by nimnim last updated on 14/Sep/20

	$If a, b and c are primitive pythagorean triples$ $then : -$ $Case (1) when a = odd,$ $b = \frac{a^{2}}{2} - \frac{1}{2} and c = \frac{a^{2}}{2} + \frac{1}{2}$ $Let a = 13, c = \frac{169}{2} + \frac{1}{2} = 85 \neq 101$ $a = 15, c = \frac{225}{2} + \frac{1}{2} = 113 \neq 101$ $Case (2) when a = even$ $b = {(\frac{a}{2})}^{2} - 1 and c = {(\frac{a}{2})}^{2} + 1$ $Let a = 20, c = {(\frac{20}{2})}^{2} + 1 = 101$ $b = {(\frac{20}{2})}^{2} - 1 = 99$ $\Rightarrow (p, q) = (20, 99) or (99, 20)$
	Answered by Rasheed.Sindhi last updated on 15/Sep/20

	$I f m & n a r e a n y p o s i t i v e$ $i n t e g e r s w i t h m > n$ $(m^{2} - n^{2}, 2 m n, m^{2} + n^{2}) i s a$ $p a t h a g o r e a n t r i p l e t .$ $T h i s r u l e i s p e r f e c t, i t c a n p r o d u c e$ $a l l t h e p a t h a g o r e a n t r i p l e t s a n d$ $t h a t a l s o m e a n s t h a t f o r e v e r y$ $t r i p l e t t h e r e e x i s t m & n w h i c h$ $c a n p r o d u c e t h a t t r i p l e t .$ $H e r e (p, q, 101) i s p a t h a g o r e a n$ $t r i p l e t s o m^{2} + n^{2} = 101$ $o r m^{2} = 101 - n^{2} . C l e a r l y$ $n ⩽ ⌊ \sqrt{101} ⌋, s o {w e}^{'} v e t o f i n d s u c h$ $n f o r w h i c h 101 - n^{2} i s p e r f e c t$ $s q u a r e b e t w e e n 1 a n d 10 .$ $H e n c e n = 1, 10 a r e o n l y p r o p e r$ $v a l u e s a n d c o r r e s p o n d i n g v a l u e s$ $o f m a r e 10, 1 . S i n c e m > n s o$ $(m, n) = (10, 1)$ $p = m^{2} - n^{2} = 10^{2} - 1^{2} = 99$ $q = 2 m n = 2 (10) (1) = 20$ $O r$ $p = 2 m n = 20$ $q = m^{2} - n^{2} = 99$ $S i n c e p, q \in Z$ $(p, q) = (20, 99), (20, - 99),$ $(- 20, 99), (- 20, - 99),$ $(99, 20), (99, - 20),$ $(- 99, 20), (- 99, - 20) .$
	Answered by 1549442205PVT last updated on 16/Sep/20
	$We known that( a,b,c )is a triple of Pythagor posirive integer numbers where b^2 +c^2 =a^2 (a,b)=1if and only if c=uv,b=((u^2 −v^2 )/2), a=((u^2 +v^2 )/2) where u,v are odd and coprime .In the our problem p^2 +q^2 =101^2 (∗) Since 101 is a prime,so (p,q)=1. Hence,p=uv,q=((u^2 −v^2 )/2),((u^2 +v^2 )/2)=101 ⇔u^2 +v^2 =202⇒v<9<u≤13 i)If u=13 then v^2 =202−169=33⇒rejected ii)If u=11 then v^2 =202−121=81=9^2 ⇒v=9⇒p=uv=99,q=((11^2 −9^2 )/2)=20 Since p,q have equal role in the equation,we obtain (p,q)∈{(99,20),(20,99).From this follows (−p,q),(p,−q),(−p,−q) also satisfying second way: If(p,q)is a root of (∗) then (−p,q),(p,−q) and (−p,−q)are also roots of (∗).Hence we just need find pairs (p,q)of positive integer numbers of the equation(∗) Since 101 is a prime number,p and q are coprime( because suppose that gcd(p,q)=d≠1 then p=md,q=nd with (m,n)=1.Then p^2 +q^2 =101^2 (∗) ⇔(m^2 +n^2 )d^2 =101^2 ⇒101⋮d which is contradiction to the fact that 101 is prime) .Furthermore,among two numbers p,q must have one number is even and other number is odd.Hence,WLOG suppose p is odd and q is even.From that we have (∗)⇔p^2 =(101+q)(101−q) Put 101+q=m,101−q=n ⇒m>n,we get p^2 =mn,q=((m−n)/2),101=((m+n)/2). We see that m and n −coprime. because of suppose the contrary, that gcd(m,n)=d≠1.Then m=m_1 d,n=n_1 d From that 101=((m+n)/2)=((m_1 +n_1 )/2)d which is impossible since 101 is prime Therefore ,m and n are coprime.From p^2 =mn and (m,n)=1 we infer m=u^2 ,n=v^2 where u and v−coprime, u>v >0 and are odd numbers.Finally, we obtain { ((p=uv(1))),((q=((u^2 −v^2 )/2)(2))),((101=((u^2 +v^2 )/2)(3))) :} (3)⇔u^2 +v^2 =202⇒v<11≤u≤13 i)For u=13⇒v^2 =202−169=33⇒∄v∈Z ii)For u=11⇒v^2 =202−121=81 ⇒v=9.Put into (1)(2) we get (p,q)∈(99,20)and due to above note and the equality of p and q in equation we get (p,q)∈{99,20),(−99,20),(20,99),(20,−99), (99,−20),(−99,−20),(−20,99),(−20,−99)}$
	$We known that (a, b, c) is a triple of$ $Pythagor posirive integer numbers$ $where b^{2} + c^{2} = a^{2}$ $(a, b) = 1 if and only if c = uv, b = \frac{u^{2} - v^{2}}{2},$ $a = \frac{u^{2} + v^{2}}{2} where u, v are odd and coprime$ $. In the our problem p^{2} + q^{2} = 101^{2} ()$ $Since 101 is a prime, so (p, q) = 1 .$ $Hence, p = uv, q = \frac{u^{2} - v^{2}}{2}, \frac{u^{2} + v^{2}}{2} = 101$ $\Leftrightarrow u^{2} + v^{2} = 202 \Rightarrow v < 9 < u ⩽ 13$ $i) If u = 13 then v^{2} = 202 - 169 = 33 \Rightarrow rejected$ $ii) If u = 11 then v^{2} = 202 - 121 = 81 = 9^{2}$ $\Rightarrow v = 9 \Rightarrow p = uv = 99, q = \frac{11^{2} - 9^{2}}{2} = 20$ $Since p, q have equal role in the$ $equation, we obtain$ $(p, q) \in {(99, 20), (20, 99) . From this$ $follows (- p, q), (p, - q), (- p, - q)$ $also satisfying$ $second way :$ $If (p, q) is a root of () then (- p, q), (p, - q)$ $and (- p, - q) are also roots of () . Hence$ $we just need find pairs (p, q) of positive$ $integer numbers of the equation ()$ $Since 101 is a prime number, p and q$ $are coprime (because suppose that$ $\gcd (p, q) = d \neq 1 then p = md, q = nd with$ $(m, n) = 1 . Then p^{2} + q^{2} = 101^{2} ()$ $\Leftrightarrow (m^{2} + n^{2}) d^{2} = 101^{2} \Rightarrow 101 ⋮ d which is$ $contradiction to the fact that 101 is prime)$ $. Furthermore, among two numbers p, q$ $must have one number is even and other$ $number is odd . Hence, WLOG suppose$ $p is odd and q is even . From that we have$ $() \Leftrightarrow p^{2} = (101 + q) (101 - q)$ $Put 101 + q = m, 101 - q = n \Rightarrow m > n, we$ $get p^{2} = mn, q = \frac{m - n}{2}, 101 = \frac{m + n}{2} .$ $We see that m and n - coprime .$ $because of suppose the contrary, that$ $\gcd (m, n) = d \neq 1 . Then m = m_{1} d, n = n_{1} d$ $From that 101 = \frac{m + n}{2} = \frac{m_{1} + n_{1}}{2} d$ $which is impossible since 101 is prime$ $Therefore, m and n are coprime . From$ $p^{2} = mn and (m, n) = 1 we infer$ $m = u^{2}, n = v^{2} where u and v - coprime,$ $u > v > 0 and are odd numbers . Finally,$ $we obtain {\begin{cases} p = uv (1) \\ q = \frac{u^{2} - v^{2}}{2} (2) \\ 101 = \frac{u^{2} + v^{2}}{2} (3) \end{cases}$ $(3) \Leftrightarrow u^{2} + v^{2} = 202 \Rightarrow v < 11 ⩽ u ⩽ 13$ $i) For u = 13 \Rightarrow v^{2} = 202 - 169 = 33 \Rightarrow ∄ v \in Z$ $ii) For u = 11 \Rightarrow v^{2} = 202 - 121 = 81$ $\Rightarrow v = 9 . Put into (1) (2) we get$ $(p, q) \in (99, 20) and due to above note and the$ $equality of p and q in equation we get$ $(p, q) \in {99, 20), (- 99, 20), (20, 99), (20, - 99),$ $(99, - 20), (- 99, - 20), (- 20, 99), (- 20, - 99)}$
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