A number theory problem by Nick Okita

Number Theory Level 5

p n + 144 = q 2 \large p^n + 144 = q^2

Let p , q p,q and n n be positive integers satisfying the equation above such that p p is a prime number . Let all the solutions of ( n , p , q ) (n,p,q) be denotes as ( n 1 , p 1 , q 1 ) , ( n 2 , p 2 , q 2 ) , … , ( n m , p m , q m ) (n_1, p_1, q_1) , (n_2, p_2, q_2) , \ldots , (n_m, p_m, q_m) . Find ∑ k = 1 m ( n k + p k + q k ) \displaystyle \sum_{k=1}^m (n_k + p_k + q_k ) .

This question was taken from IME (Instituto Militar de Engenharia) 2009-2010 entrance exam.


The answer is 72.

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Nick Okita by Nick Okita , 23, Brazil

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1 solution

Adarsh Kumar
Sep 15, 2014

W e c a n w r i t e t h e g i v e n e q u a t i o n a s : We\ can\ write\ the\ given\ equation\ as: p n + 1 2 2 = q 2 p^{n}+12^{2}=q^{2} T h i s i s b a s i c a l l y a p y t h a g o r e a n t r i p l e t . This\ is\ basically\ a\ pythagorean\ triplet. T h u s , w e h a v e p n = x 2 f o r a n y p o s i t i v e i n t e g e r ′ x ′ Thus,we\ have\ p^{n}=x^{2}\ for\ any\ positive\ integer\ 'x' ⟹ x 2 + 1 2 2 = q 2 \Longrightarrow\ x^{2}+12^{2}=q^{2} ⟹ 144 = q 2 − x 2 \Longrightarrow\ 144=q^{2}-x^{2} ⟹ 144 = ( q − x ) ( q + x ) \Longrightarrow\ 144=(q-x)(q+x) N o w , j u s t f a c t o r i s e 144 a n d y o u w i l l g e t t h e v a l u e s . Now,just\ factorise\ 144\ and\ you\ will\ get\ the\ values. K e e p i n m i n d t h a t ′ p ′ s h o u l d b e a p r i m e . Keep\ in\ mind\ that\ 'p'\ should\ be\ a\ prime.

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p n = q 2 − 12 2 {p}^{n} = {q}^{2} - {12}^{2}

p n = ( q − 12 ) ( q + 12 ) {p}^{n} = (q-12)(q+12)

Therefore, q − 12 = p a q-12 = {p}^{a}

q + 12 = p b q+12 = {p}^{b}

Subtracting the above 2 equations,

2 ∗ 12 = p b − p a 2*12 = {p}^{b} - {p}^{a}

Case 1 When a = 0

24 = p b − 1 24 = {p}^{b} - 1

p b = 25 {p}^{b} = 25 . Clearly, p = 5, b = 2

Case 2 When a >= 1

2 3 ∗ 3 = p a ( p l − 1 ) {2}^{3}*3 = {p}^{a}({p}^{l} - 1) [such that a + l = b, and a<b(which is obvious)]

Therefore, p a = 2 3 o r 3 1 {p}^{a} = {2}^{3} or {3}^{1}

By further calculations for finding n for each p, we can see that

p = 5, n = 2, q = 13

p = 2, n = 8, q = 20

p = 3, n = 4, q = 15

Kartik Sharma - 6 years, 7 months ago

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Did the same

Aditya Kumar - 5 years ago

I am not entirely correct because I said that it would be a pythagorean triplet but that is not necessary. eg; 3 3 + 1 3 2 = 1 4 2 3^{3}+13^{2}=14^{2}

Adarsh Kumar - 6 years, 9 months ago

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Then, change your solution!

Kartik Sharma - 6 years, 7 months ago

Did I already see this problem on this site?

Krishna Ar - 6 years, 9 months ago

@Krishna Ar

Adarsh Kumar - 6 years, 8 months ago

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What? Why did you mention me? :O

Krishna Ar - 6 years, 8 months ago

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Look at my comment about pythagorean triplets.

Adarsh Kumar - 6 years, 8 months ago

@Satvik Golechha please see my comment below about pythagorean triplets!!!

Adarsh Kumar - 6 years, 8 months ago

@Calvin Lin please see my comment below about pythagorean triplets

Adarsh Kumar - 6 years, 8 months ago

well n does not limit to a perfect square, i did it by trial and error

akash deep - 6 years, 8 months ago

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