Sum of Some Squared Divisors

First of all we will decide the parity of $N$ which will be the most basic idea to find $d_{2}$ .

For the sake of contradiction,suppose $N$ is odd,which implies that all of its divisors are odd. So,all $d_{1},d_{2},d_{3},d_{4}$ are odd and hence their squares are odd too. But, sum of $4$ odd numbers is always even,which implies $N$ is even which disturbs our assumption. So, $d_{2}=2.$ $\Rightarrow N=5+d_{3}^{2} + d_{4}^{2}.$

Again for the sake of contradiction,suppose

$(i)$ Both $d_{3}$ and $d_{4}$ are odd $\Rightarrow d_{3}^{2},d_{4}^{2} \equiv 1(mod4), \Rightarrow d_{3}^{2}+d_{4}^{2}+5 \equiv 7 (mod4), \Rightarrow N \equiv 3 (mod4),\Rightarrow N \equiv 1(mod2)$ Contradiction!, because as $d_{2}=2$ , $N$ is even. By the same method we can that both $d_{3}$ and $d_{4}$ cannot be both even.

The above proof shows that one of $d_{3}$ and $d_{4}$ is even and the other one is odd. Suppose $d_{3}$ is even,then $d_{3}=4.$ So, $d_{4}$ is an odd prime. $N=21+d_{4}^{2} \equiv 2 (mod4)$ ,but $N$ is a multiple of 4.So $d_{3}$ is odd and $d_{4}$ is even.

As $d_{3} > d_{2}$ , $d_{3} = p$ ,a prime.And as $d_{4}$ is even,it can be either a power of $2$ or $2p$ .Suppose $d_{4}$ is a power of $2.$ Again by checking modulo 4, we get a contradiction. So $d_{4} = 2p.$ $\Rightarrow N=5+5p^{2} \equiv 0 (mod5)$ $\Rightarrow p=5$ So we get $d_{1}=1,d_{2}=2,d_{3}=5,d_{4}=10$ So $N=\boxed{130}.$

If $N$ is odd, then $d_1,d_2,d_3,d_4$ are all odd and $N = d_1^2+d_2^2+d_3^2+d_4^2$ will be even which is a contradiction. Hence, $N$ is even and $d_2=2$ . Now, for $N$ to be even, exactly one of $d_3$ and $d_4$ must be even.

Since exactly two of $d_i$ must be odd, and odd squares give a remainder of 1 when divided by 4, thus $N = d_1^2+d_2^2+d_3^2+d_4^2 \equiv 1+0+1+0 \equiv 2 \pmod{4}$ . Therefore, $N$ is not divisible by 4. [Note: This step is often missed, and it's very important later that $4 \not \mid N$ . - Calvin]

Let the second smallest even factor of $N$ be $2k (k>1)$ . Since $4$ is not a factor of $N$ , $k>2$ . Since $k \mid N \Rightarrow d_3 = k$ and $d_4 = 2k$ since they must be the two smallest factors larger than 2. [Note: This is the step that is most often missed by students, who argue by 2 odd, 2 even case. - Calvin] Therefore, $N = 1^2+2^2+k^2+(2k)^2 = 5 + 5k^2 = 5(1+k^2) \Rightarrow 5 \mid N$ .

If $d_3=3$ then $d_4=6$ and $N = 1^2 + 2^2 + 3^2 + 6^2 = 50$ which contradicts the fact that $6$ is the 4th smallest factor since that should clearly be 5. [Note: This step was also often missed by students who said that since $5 \mid N$ , hence $d_3 = 5$ . That need not be true, especially if they didn't state that $4 \not \mid N$ , since we then could have $d_1 =1, d_2=2, d_3=3, d_4=4$ as a missing case. - Calvin]

Therefore since $3 \nmid N$ and $4 \nmid N$ , $d_3$ must be $5$ since we have shown that $N$ must be divisible by $5$ . Hence, $d_4=10$ and the only solution for $N$ is $N = 1^2 + 2^2 + 5^2 + 10^2 = 130$ . A quick check shows that this $N$ satisfies the conditions.

If $d_2 \neq 2$ , then all factors of $N$ are odd, so that $d_1,d_2,d_3,d_4$ are all odd. But that would make $N = d_1^2+d_2^2+d_3^2+d_4^2$ even! Thus we deduce that $d_2=2$ .

1. If $d_3=3$ then $N = 14 + d_4^2$ , and so $d_4|14$ . Since $N$ is even, $d_4$ must be even, and hence $d_4=14$ . But that makes $N=210$ , which has $5$ as a factor, so $d_4=14$ is not possible.

2. If $d_3=4$ then $N = 21 + d_4^2$ , and so $d_4|21$ . Thus $d_4$ is odd, and hence $N \equiv 2$ modulo $4$ , so is not divisible by $4$ . This case is not possible either.

3. Trying $d_3=5$ , we need $N = 30 + d_4^2$ and $d_4|30$ . Since $3$ is not a factor of $N$ we deduce that $d_4=10$ , and hence $N = 130$ . The factors of $130$ are $1,2,5,10,13,26,65,130$ .

Disclaimer: This solution is by no means a "here is the answer to everything you ever wanted to know about this question," rather it is a step-by-step playback of what went through in my mind. Take it as you will.

First, $2$ must be a factor of $N$ . If $N$ was odd, then its divisors would be all odd, so we would have the sum of four odd numbers constituting $N$ , making $N$ even. Thus, the equation turns into

$N = 5 + {d_3}^2 + {d_4}^2$

My motivation for doing this is that the problem becomes so much simpler if this is true, and it is quite an easy statement to prove. Let's try to figure out more about these numbers. Well, we can't have $4$ as a factor of $N$ , because then four must be included in the sum as well, creating $21 + d^2 = N$ , with $4 | N$ . So $d$ must be odd, but $(2n-1)^2 \equiv 1 \pmod 4$ , so then $N \equiv 2 \pmod 4$ , a contradiction to our divisibility statement. So $N$ has only one factor of $2$ .

Now that we've figured that out, let's experiment with the idea that $d_4 = 2 * d_3$ . Then our equation is $5 + 5{d_3}^2 = N$ . Thus, $N$ must be divisible by $5$ . In addition, $d_3$ cannot equal $3$ because then $3$ not divide $N$ , thus in this example $d_3$ must equal $5$ . Smells like this is a good solution -- let's just go ahead and check. $5 + 125 = 130 = N$ . $N = 2 \times 5 \times 13$ , so $d_1 = 1, d_2 = 2, d_3 = 5, d_4 = 10$ , just as we used. So aha! $N = 130$ .

If all four divisors are odd, then N is even because the sum of four odds is even (they remain odd when squared). But then it has 2 as a divisor, and it would be second on the list. So not all four divisors are odd. Therefore N has an even divisor, and 2 in particular. So the first two divisors are 1 and 2.
By checking d3=3,4,5, one finds the solution 1,2,5,10.

Consider 130. Its divisors are 1,2,5,10,13,65,130. Aranging in ascending order as 1=1<2<5<10<13<65<130=130. Now we have 1^2+2^2+5^2+10^2=1+4+25+100=130. This satisfy the require condition. Hence the solution.

We will utilize $\pmod{2}$ , $\pmod{4}$ , and a simple contradiction argument to show that $N=\boxed{130}$ .

Lemma: $d_2=2$ .

Proof: Obviously $2|N\iff d_2=2$ . Now suppose $N$ is odd. Then all of its divisors are also odd. But the sum of four odd numbers is even, contradiction! So $2|N$ and $d_2=2.\blacksquare$

Our condition now reduces to $N=5+d_3^2+d_4^2$ . We note that since $N$ is even, exactly one of $d_3, d_4$ is odd. Then taking $\pmod{4}$ , we see that $N\equiv 1+1\equiv 2 \pmod{4}$ . Thus, $2|N$ but $4\nmid N$ . Then, the even divisor $d_x$ is of the form $2a$ , where $a$ is odd. But then $a|N$ and $a<2a$ , so we see that $d_3=a$ and $d_4=2d_3$ . Now, our condition simplifies to $N=5(d_3^2+1)$ . So $5|N$ , and we easily see that $d_3=5\implies d_4=10$ . This gives the solution $\{d_1,d_2,d_3,d_4,N\}=\{1,2,5,10,130\}$ , which we see works. Thus, $N=\boxed{130}$ .

Motivation: N has prime divisors, and the smallest must be $d_2$ . What can we say about $d_3$ and $d_4$ ?

First, note that 2 divides N: Otherwise, N is a sum of 4 odd squares, so N is even, a contradiction. Thus $d_2=2$ .

Second, note that 4 does not divide N: Taking the sum mod 4 gives $N=1+4+d_3^2+d_4^2 \equiv 1+d_3^2+d_4^2 \equiv 1+0, 1,$ or $2 \equiv 1, 2,$ or $3$ , a contradiction.

Third, note that $d_4=2d_3$ . This is because N is even, and the first two terms sum to an odd number, so the second two must as well. If $d_3$ were even, then it would be 2*k for some odd k, and since k can't be one, k has to have an odd prime factor, which is smaller and must also divide N, a contradiction. Let $d_3=p$ . Then 2p divides 2k divides N, so 2p divides N. Also, by the similar logic as before, any other even divisor must be larger; thus, since $d_3$ odd implies $d_4$ is even, $d_4=2p=2d_3$ .

From this, we get $N=1+4+p^2+(2p)^2=(1+4)(1+p^2)=5(1+p^2)$ . Since 5 divides N, 3 can't divide N, since 2*3>5; thus 5=p, and $N=1+4+25+100=130$

Let S be the set {d1, d2, d3, d4}

First, we know that d1=1. (it was given in the problem statement)

Next, determine if N is either odd or even. Say N is odd. Then either 1 or 3 of the elements of S must be odd. But there are 4 elements in S. Thus, it must contain at least one element that is even. And if a number has a factor which is an even number, then the number must be an even number. And so we can now say that N is even. Without further thinking, we are now certain that d2=2.

Now we know that N is even, then we can conclude that S has 1 more element which is an odd number and another which is even. For the odd number element, must certainly be a prime or else there will be some other factor of N which is an odd prime that is lower than the said element. Let this element be named p. For the even number element, it must certainly be of the form 2x, where x is a prime (since if it is not, then there must be another number of lower value which is a factor of N)

As of now, we have the following equation for N: N = (1^2) + (2^2) + (p^2) + ((2x)^2)

We have two cases for x:

case 1:

x=2
N = 1 + 4 + (p^2) + 16 = (p^2) + 21
also, we can say that N = 4*k*p, where k is some other integer which needs to be multiplies to 4*p in order to form N.

N = (p^2) + 21 = 4 k p -> (p^2) - 4 k p + 21 = 0 we try to factor this and obtain integer values for both k and p and also a prime number for p. the only seemingly possible factorization then would be N = (p-3) * (p-7) = (p^2) - 10p + 21 but would yield: 4*k = 10 -> k = 2.5 which is not valid based on what we want. thus, x must not be equal to 2

case 2:

x is a prime.

obviously, if x is a prime, then it must be p (the other element of S) and so we have: N = 1 + 4 + (p^2) + (2p)^2 = 5 + (p^2) + 4(p^2) = 5 + 5(p^2) = 5 (1 + (p^2)) N = 5 (1 + (p^2)) needless to say, we found out that 5 is a factor of N. thus, d3 = 5. and, d4 = 2p = 2*5 = 10.

finally, N = (1^2) + (2^2) + (5^2) + (10^2) = 1 + 4 + 25 + 100 = 130!

if $N$ is Odd , $d_2,d_3,d_4$ will be odd. that's contradition because $N=d_1+d_2+d_3+d_4$ isn't odd.

so $N$ is Even and $d_2=2$ and we get new equation that is

$N=1^2+2^2+d_3^2+d_4^2 \rightarrow N-d_3^2 = 5+d_4^2$

to observe LHS that divides by $d_3$ . so $d_3|(d_4^2+5)$ and on the other hand $d_4|(d_3^2+5)$

if $d_3$ is Even , $d_3$ can be only $2^2$ and $d_4=7,21$ . but that's contradiction. so $d_3$ is ood and must be prime number. that makes $d_4$ must be Even.

Let $d_3=p$ , we will get $d_4=2p$ . from $d_3|(5+d_4^2)$ we know that $d_3=5$

therefore $N=130$ .

suppose N is odd then all the factors of N would be odd so d1^2+d2^2+d3^2+d4^2 is even, contradiction.

Now N is even, and divisible by 4

d1=1 and d2=2. So 1+4+d3^2+d4^2 =N=0 (mod 4)

d3^2+d4^2=3 (mod 4) which is not possible.

from above argument, one of the d3 or d4 must be even of the form of 2p where p is prime. and p doesn't equal to 2.

but p < 2p so d4=2p and d3=p.

hence 1 +4+p^2+4p^2=5(1+p^2)=N

5 devides N. now p=3 is not possible so p must be 5 because d3 is third smallest divisor.

hence N=130.

Assume N is odd then d1, d2, d3, d4 must be odd too. The sum of them must be even. so d1=1, d2=2, d3 and d4 is <odd, even>. The sum of ODD^2+EVEN^2 = 1(mod 4), so N=2(mod 4), then d4=2 d3. N=5+5 d3^2. d3=5. N=130.