N is not that big!

Number Theory Level 5

Find the largest integral n n such that 12 n − 119 12n-119 and 75 n − 539 75n-539 are both perfect squares.


The answer is 20.

Ojas Singh Malhi by Ojas Singh Malhi , 18, India

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1 solution

Ojas Singh Malhi
Sep 13, 2017

Observe that 25 ( 12 n − 119 ) − 4 ( 75 n − 539 ) = − 819 25(12n-119)-4(75n-539)=-819
Let 12 n − 119 = x 2 12n-119=x^2 and 75 n − 539 = y 2 75n-539=y^2 ,
We get, 4 y 2 − 25 x 2 = 819 4y^2-25x^2=819
= > ( 2 y ) 2 − ( 5 x ) 2 = 819 =>(2y)^2-(5x)^2=819
=> ( 2 y − 5 x ) ( 2 y + 5 x ) = 819 (2y-5x)(2y+5x)=819
Since, the difference of the two factors is divisible by 10 10
Therefore, there are only 2 satisfactory factorizations of 819 i.e. 63 ∗ 13 63*13 and 117 ∗ 7 117*7
Comparing, we get 2 y − 5 x = 13 2y-5x=13 and 2 y + 5 x = 63 2y+5x=63
=> 4 y = 76 4y=76
=> y = 19 y=19 and x = 5 x=5
OR
2 y − 5 x = 7 2y-5x=7 and 2 y + 5 x = 117 2y+5x=117
=> 4 y = 124 4y=124
=> y = 31 y=31 and x = 11 x=11
Consequently, n = 12 n=12 and n = 20 n=20
20 > 12 20>12
=> n = 20 \boxed{n=20}


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Pi Han Goh - 3 years, 8 months ago

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Thanks, it worked!

Ojas Singh Malhi - 3 years, 8 months ago

You missed 3 ⋅ 273 3 \cdot 273 as a possible factorization, which leads to y = 69 , x = 27. y=69,x=27. Unfortunately, in this case, n = 212 / 3 n=212/3 is not an integer.

Patrick Corn - 3 years, 7 months ago

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