Sum of squares! Ha!

Number Theory Level 4

Find all integers a ∈ Z a \in \mathbb{Z} such that a 2 + 4 2 a + 1 \large{\dfrac{a^2 +4}{2a + 1}} is also an integer.

Input your answer as the sum of the squares of all possible values of a \boxed{a} .


The answer is 146.

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Sualeh Asif by Sualeh Asif , 21, Pakistan

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2 solutions

First start with dividing polynomial

a 2 + 4 2 a + 1 = a 2 − 1 4 + 17 / 4 2 a + 1 \displaystyle \frac{a^{2}+4}{2a+1}=\frac{a}{2}-\frac{1}{4}+\frac{17/4}{2a+1}

let's this fraction to be an integer

a 2 + 4 2 a + 1 = n = a 2 − 1 4 + 17 / 4 2 a + 1 \displaystyle \frac{a^{2}+4}{2a+1}=n=\frac{a}{2}-\frac{1}{4}+\frac{17/4}{2a+1}

4 n = 2 a − 1 + 17 2 a + 1 \displaystyle 4n=2a-1+\frac{17}{2a+1}

hence this expression will be an integer if 17 is divisible by 2 a + 1 2a+1

So 2 a + 1 = ± 1 , ± 17 \displaystyle 2a+1=\pm 1,\pm 17

a = 0 , − 1 , 8 , − 9 a=0,-1,8,-9

0 2 + ( − 1 ) 2 + 8 2 + ( − 9 ) 2 = 146 0^{2}+(-1)^{2}+8^{2}+(-9)^{2}=\boxed{146}

15 Helpful 0 Interesting 1 Brilliant 0 Confused

Great use of the Dividing Polynomial! @คลุง แจ็ค Upvoted!

P.s. For a better solution do add in how you used the dividing polynomial

Sualeh Asif - 5 years, 11 months ago

Nice. Up voted. Simple approach.

Niranjan Khanderia - 5 years, 11 months ago
Kazem Sepehrinia
Jun 27, 2015

2 a + 1 ∣ a ( 2 a + 1 ) − 2 ( a 2 + 4 ) = a − 8 2 a + 1 ∣ 2 a + 1 − 2 ( a − 8 ) = 17 2 a + 1 = ± 1 , ± 17 2a+1| a(2a+1)-2(a^2+4)=a-8 \\ 2a+1| 2a+1-2(a-8)=17 \\ 2a+1=\pm1, \pm 17 So, a = − 9 , − 1 , 0 , 8 a=-9, -1, 0, 8 .

13 Helpful 0 Interesting 0 Brilliant 0 Confused

Very good approach. Up voted.. I learn a new method today. Thanks. Probably you may explain this method in detail for others to understand. I took time to understand .

Niranjan Khanderia - 5 years, 11 months ago

Can you explain this in more detail?

Shashank Rammoorthy - 5 years, 11 months ago

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2 a + 1 ∣ a ( 2 a + 1 ) − 2 ( a 2 + 4 ) = a − 8 2 a + 1 ∣ 2 a + 1 − 2 ( a − 8 ) = 17 2 a + 1 = ± 1 , ± 17 Multiple of(2a+1) is used to eliminate a 2 f r o m ( a 2 + 4 ) . I think this may be the normal way in Number Theory 2a+1| a(2a+1)-2(a^2+4)=a-8 \\ 2a+1| 2a+1-2(a-8)=17 \\ 2a+1=\pm1, \pm 17\\\text{Multiple of(2a+1) is used to eliminate } a^2~ from ~ (a^2+4).\\\text{I think this may be the normal way in Number Theory}

Niranjan Khanderia - 5 years, 11 months ago

Hi shashank,

If integer x x is a common divisor of y y and z z , then it divides every linear combination of y y and z z . In other words if x ∣ y x|y and x ∣ z x|z then x ∣ m y + n z x|my+nz for arbitrary integers m m and n n .

Now for this problem let x = 2 a + 1 x=2a+1 , now we must have 2 a + 1 ∣ a 2 + 4 2a+1|a^2+4 , so let y = a 2 + 4 y=a^2+4 and 2 a + 1 ∣ 2 a + 1 2a+1|2a+1 , so let z = 2 a + 1 z=2a+1 . I choose m m and n n in a way that degree of a a decreases on the RHS of relations. I repeat this procedure to reach a number on the RHS. Rest is easy. I Hope I'm clear.

Kazem Sepehrinia - 5 years, 10 months ago

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sir how do u know that we have to write a 2 + 4 2 a + 1 a s a 2 − 1 4 + 17 / 4 2 a + 1 \displaystyle \frac{a^{2}+4}{2a+1}as\frac{a}{2}-\frac{1}{4}+\frac{17/4}{2a+1} @Kazem Sepehrinia

sakshi rathore - 5 years, 10 months ago

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@Sakshi Rathore – Hi,

I think you should ask @คลุง แจ็ค

Kazem Sepehrinia - 5 years, 10 months ago

@Sakshi Rathore – a 2 + 4 = 1 2 ∗ ( 2 a 2 + a − a ) + 4 = 1 2 ∗ ( 2 a 2 + a ) − 1 2 ∗ 1 2 ( 2 a + 1 − 1 ) + 4 = 1 2 ∗ ( 2 a 2 + a ) − 1 2 ∗ 1 2 ∗ ( 2 a + 1 ) − 1 2 ∗ 1 2 ∗ ( − 1 ) + 4 = 1 2 ∗ a ∗ ( 2 a + 1 ) − 1 4 ∗ ( 2 a + 1 ) + 1 4 + 4 ∴ a 2 + 4 2 a + 1 = 1 2 ∗ a ∗ ( 2 a + 1 ) − 1 4 ∗ ( 2 a + 1 ) + 1 4 + 4 2 a + 1 = 1 2 ∗ a − 1 4 + 17 / 4 2 a + 1 a^2+4=\dfrac 1 2 *(2a^2+a -a)+4\\ =\dfrac 1 2 *(2a^2+a) -\dfrac 1 2 *\frac 1 2(2a+1-1)+4\\ =\dfrac 1 2 *(2a^2+a) -\dfrac 1 2 *\frac 1 2 *(2a+1)-\dfrac 1 2 *\frac 1 2 *(-1)+4\\ =\dfrac 1 2 *a*(2a+1) -\dfrac 1 4 *(2a+1)+\dfrac 1 4+4\\ \therefore~\dfrac{a^2+4}{2a+1}\\ =\large \dfrac {\frac 1 2 *a*(2a+1) -\frac 1 4 *(2a+1)+\frac 1 4+4}{2a+1}\\ =\dfrac 1 2 *a-\dfrac 1 4 +\dfrac{17/4}{2a+1}

Niranjan Khanderia - 5 years, 10 months ago

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