Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How many integral solutions (x, y) exist satisfying the equation |y| + |x| ≤ 4

My approach:

I have made the graph after opening the the modulus in the above equation by making four equations.

Now it is a square with co-ordinates (4,0)(0,4)(-4,0)(0,-4).

Now I am stuck and don't know how to calculate integral solutions. It should be integral boundary points plus the integral points inside the area.

I know about the Pick's theorem in which we can find the integral points by using area and boundary points but I need to know how to calculate the integral points without it.

Answer is 41. In my book it is given as 9+2(7+5+3+1)=41 [which I am not getting]

Kindly help in solving the same.

share|cite|improve this question
Is it perhaps $\le 4$ instead of $=4$ ? – Dietrich Burde Aug 9 '13 at 18:10
Consider the ways to partition each of $0,1,2,3,4$ into sums of two positive integers... then consider that e.g. $a+b=c$ yields the solutions $(a,b),(a,-b),(-a,b),(-a,-b)$ – oldrinb Aug 9 '13 at 18:19

Clearly, $0\le|x|\le4\implies -4\le x\le 4 $

If we need to find the number of integral points inside the area $|x|+|y|=4$

For $0\le a\le4,$ if $x= \pm a,|x|=a,|y|\le 4-a\iff -(4-a)\le x\le 4-a,$ so $x$ can assume $2(4-a)+1=9-2a$ values including $0$

If $a=0,a=-a,x$ can assume $2(4-0)+1=9$ values

So, the number integer points will be $9+2\sum_{1\le r\le 4}(9-2a),$ the multiplier $2$ is due to the fact that there is one $-a$ for each integer $a\in[1,4]$

share|cite|improve this answer

We begin by counting the lattice points satisfying $x+y\leq4$ in the first quadrant. Collect the points lying on a line $x+y={\rm const.}$ into a group. In this way we obtain $1+2+\ldots+5={5\cdot 6\over 2}=15$ points. Now there are four such triangles, makes $60$ points. In this way we have counted the points $\ne(0,0)$ on the axes twice and the origin four times. Therefore we deduct $4\cdot4$ and $3$, leaving a final $41$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.