Find the value of a

Number Theory Level 3

Let $n$ be a positive integer such that $n^{2} + 35n + 174$ is a perfect square $a^{2}$ . Find the positive value of $a$ .

The answer is 132.

2 solutions

Chang Jia Geng
Dec 22, 2015

Intended solution: After factorisation of the polynomial, we see that the difference betwen two squares x^2 and y^2 is 23, so (x+y)(x-y) = 23. Solving for integer roots we obtain x=12 and y=11. Hence a = x×y = 132

Azzim Habibie
Dec 21, 2015

n^2 + 35n + 174 = a^2 Factorize LHS : ( n + 6 )( n + 29 ) = a^2 RHS is a square number, then LHS must be a square number too. Then let n + 6 = x^2 and n + 29 = y^2
n + 6 = x^2 》n = x^2 - 6 n + 29 = y^2 》n = y^2 - 29 Then: x^2 - 6 = y^2 - 29 y^2 - x^2 = 23 ( x - y )( x + y ) = 1 . 23 x - y = 1.......(1) x + y = 23...(2) Sum (1) and (2), then we got : x = 11 y = 12

Then n = 11^2 - 6 = 115 So, (115 + 6 )( 115+ 29 ) = 121 . 144 = (11 . 12)^2 = a^2 》 a = 11 . 12 = 132.

As you mentioned that if $(n+6)(n+29)$ is a perfect square then $(n+6)$ and $(n+29)$ must also be perfect squares.Is it always true? Consider $(n+1)(n+7)$ is a perfect square for $n=1$
but $(n+1)$ and $(n+7)$ are not separately perfect squares.Sorry if I am missing something:)

Siddharth Singh - 5 years, 5 months ago

Great example! I think it only works if the greatest common divisor between the two terms is 1 i.e. gcd(121,144) = 1. He first assumed this was the case, then verified it. In your example gcd(2,8)=2, and only the unique pair (1,16) works.

Chang Jia Geng - 5 years, 5 months ago

Find the value of a

The answer is 132.

2 solutions

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