Avi's integers

If it is both 2013 and 2045 more than a square, the squares must differ by 2045-2013=32. $m^2-n^2=32$ $(m-n)(m+n)=32$ Since $(m-n)+(m+n)=2m$ , the sum of the factors must be even, and so $m-n=2,4$ $m+n=8,16$ Solving, we get $(m,n)=(9,7),(6,2)$ . Then, the sum of the numbers is $2045+2^2+2045+7^2=4143$ The answer is $\boxed{143}$ .

We let our integers be equal to $x$ , and any two integers be $n,k$ . Now, we write $\begin{aligned} x &= n^2 + 2013 \\ x &= k^2 + 2045 \end{aligned}$ Or, after setting the two equations equal and some rearranging, $(n+k)(n-k) = 32$ . We note that for $n,k$ to be integers, the parity of $n+k$ and $n-k$ have to be the same. We also disregard the fact that $n,k$ can be different signs because they are squared. Listing out the factors of $32$ , we can find the two systems $\begin{aligned} n + k &= 16 \\ n - k &= 2. \end{aligned}$ and $\begin{aligned} n + k &= 8 \\ n - k &= 4 \end{aligned}$ which yields the solutions $(n.k) = (9,7), (6,2)$ . Plugging these back into our first equation, we see that the sum of the last three digits of all possibilities of $x$ are $13+81+13+36$ , which yields the desired answer, $\boxed{143}$

Let $n$ be one of the integers we want to find, and let $n = x^2+2013$ and $n = y^2 + 2045$ for nonnegative integers $x$ and $y.$ Then we can subtract these two equations to get $0 = x^2-y^2 -32,$ or $32 = x^2-y^2.$ Notice that we can factor the right-hand side to $32 = (x-y)(x+y).$ We can now test all possible cases to find all possible values of $n.$ To make it easier, we make the following points:

• Note that $x-y$ and $x+y$ have the same parity, since their difference is $2y,$ an even number. Thus, since their product is even, both $x-y$ and $x+y$ are even.

• Note that $x+y$ is positive, so $x-y$ must also be positive.

• Note that $x+y > x-y,$ since $y > 0.$

Therefore, we have only the two cases $\begin{cases} x+y = 16 \\ x-y = 2 \end{cases}$ and $\begin{cases} x+y = 8 \\ x-y = 4 \end{cases}.$ We can then easily solve these systems to find that $(x,y) = (9, 7), (6, 2).$ Therefore, $n = 6^2 + 2013 = 2049$ or $n = 9^2 + 2013 = 2094.$ The answer is $2049 + 2094 = 4143 \rightarrow \boxed{143}.$

The sum of a perfect square $n^{2}$ and 2045 is 2013 more than another perfect square $m^{2}$ :

$n^{2} + 2045 = m^{2} + 2013$

$n^{2} + 32 = m^{2}$

$32 = m^{2} - n^{2}$ ;

$m^{2} - n^{2}$ is a difference of squares and can factor out to:

$32 = (m - n)(m + n)$

The possible factors of 32 are $1 \times 32$ , $2 \times 16$ , and $4 \times 8$ without repeating terms.

Case 1: $(m - n) = 1$ and $(m + n) = 32$ Solving for m: $2m = 33$ m is not an integer and so this case fails

Case 2: $(m - n) = 2$ and $(m + n) = 16$ Solving for m: $18 = 2m, m = 9$ and so $n = 7$ and a possible number is $7^{2} + 2045 = 2094$

Case 2: $(m - n) = 4$ and $(m + n) = 8$ Solving for m: $12 = 2m, m = 6$ and so $n = 2$ and a possible number is $2^{2} + 2045 = 2049$

$2094 + 2049 = 4143$ and so the last three digits are $143$

Let $n$ be a number that has the proprety in the text. Then there exist $a,b$ positive integers such that $n=a^2+2013=b^2+2045$ Rearranging and factoring we get $(a+b)(a-b)=32=2^5$ Now, $a+b>a-b$ and so we have the following possible couples: $(a+b,a-b)=(32,1),(16,2),(8,4)$ from which we get $(a,b)=(9,7),(6,2)$ which yelds $n=a^2+2013=9^2+2013$ or $n=a^2+2013=6^2+2013$ And the sum is $9^2+2013+6^2+2013=4143$

Let $n$ be the required number.From the question, we get that both $n-2013$ and $n-2045$ are square of integers. So let $n-2013=p^2$ and $n-2045 =q^2$ . Subtracting we get $p^2-q^2 =32$ $\Rightarrow (p+q)(p-q) = 32$ . And the positive factors of $32$ are $1,2,4,8,16,32$ . Note * we took only the positive values of $32$ because negative and positive both values of p and q will yield the same value of $n$ . Also we will assume that $p \geq q$ so that we do not get negative value of $p$ or $q$ . So the following cases arises: * CASE 1: * When $(p+q)(p-q)=1 \times 32 \Rightarrow p-q=1$ and $p+q=32.$ But this doesnot give integer values of p and q. * CASE 2 * When $p-q=2 and p+q=16$ Thus we get $p=9$ and $q=7$ implying * n= 2094 *. * CASE 3: * When $p-q = 4 and p+q=8$ Thus we get $p=6$ and $q =2$ implying * n=2049 . Therefore we get two values of $n$ whose sum is 4143 . Hence the required answer is * 143 *

We can see that:

$(n+k)^2-n^2=2kn+k^2$

That shows that the difference between two squares with k between their roots is $2kn+k^2$ if n is the root of the smaller square. Note that $2kn$ is always even and k^2 is odd if k is odd and even if k is even. Also note that $2045-2013=32$ so we can choose k=2 and solve for n:

$2 \times 2n+2^2=32 \iff n=7$

Then we can see that $49+2045=2013+81=2094$ . We choose k=4 (k can't be even because 32 is even) and solve for n:

$2 \times 4n+4^2=32 \iff n=2$

Then we can see that $4+2045=36+2013=2049$ . If k>5 we would get n<0 but that doesn't give a valid solution. Hence the answer is $2094+2049=4143$ .