2011

Algebra Level pending

$X^2 + 2011 = Y^2$ where X and Y are both positive integers. Find value of 3X + Y

The answer is 4021.

2 solutions

Abdelhamid Saadi
Jul 9, 2016

$(Y - X)( Y + X) = 2011$

Since 2011 is prime, we have: $Y - X = 1 \quad and \quad Y + X = 2011$ Then $Y = 1006 \quad and \quad X = 1005$

Dominick Hing
Oct 7, 2014

${ 1 }^{ 2 }+3={ 2 }^{ 2 }$

${ 2 }^{ 2 }+5={ 3 }^{ 2 }$

${ 3 }^{ 2 }+7={ 4}^{ 2 }$

...

${ X }^{ 2 }+2011={ Y }^{ 2 }$

From the pattern we can clearly tell that $2011=2Y-1$ .

Thus $2Y=2012$ and $Y=1006$ . From the pattern, $X=Y-1$ so

$X = 1005$

$3X+Y = 3\times 1005+1006=4021$

a little more care should be taken... for example,

$1^2+15=4^2$

$2^2+21=5^2$

etc.

In general, you could write $x^2+2011=(x+a)^2$ for some integer $a$ . Expanding, you get $2011=a(2x+a)$ . Since 2011 is prime, we must have $a=1$ and $2x+a=2011$ , which gives $x=1005, y=1006$ , and $3x+y=4021$ .

Anthony Kirckof - 5 years, 4 months ago

Although this wasn't so right to solve, but this pattern actually came in my mind and helped me solved this question. I observed the pattern of the difference in square of two subsequence numbers is the sum of the two subsequence number when I was still small.

AhGuek Duck - 4 years, 5 months ago

2011

The answer is 4021.

2 solutions

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