Pairs of Perfect Squares

Since this question requires that when the digits of $N^2$ are increased by 3 , the result is also a perfect square, it is possible to solve this case by case using the difference of squares formula. Let $N$ be the original number, and let $X^2$ be the result after adding 3 to each digit of $N^2$ . Since $N \leq 100$ , $N^2$ has a maximum of 5 digits. Thus, there are 5 cases to consider:

Case 1: $N^2$ has 1 digit. We see that $(X-N)(X+N)=3$ . There is only 1 possible factorisation, namely $1\times3=3$ . N is thus $\frac {(X+N)-(X-N)}{2} = \frac {3-1}{2}=1$ . In this first case, $1\leq N<4$ . If not, $N^2$ will not have only 1 digit. Since N=1 fits within these bounds, it is a possible value of N. Upon checking, we see that $1^2+3=4=2^2$ $N=1$ .

Case 2: $N^2$ has 2 digits. There are 2 possible factorisations of $33$ : $1 \times 33$ and $3 \times 11$ . In this case, $4 \leq N < 10$ , or $N^2$ will not have exactly 2 digits. In the first factorisation, $\frac {33-1}{2}=16 \nless 10$ , thus $N=16$ is not a possible solution. In the second factorisation, $\frac {11-3}{2}=4$ , and $4 \leq 4 <10$ , so $N=4$ is a possible value. $N=4$ .

Case 3: $N^2$ has 3 digits. We follow similar steps to find 3 possible 2-term factiorisations of $333$ : $1 \times 333, 3 \times 111$ and $9 \times 37$ . Upon checking, we find that the first two factorisations result in $N^2$ having more than 3 digits, leaving only the last factorisation. However, even though $\frac {37-9}{2}=14 < 32$ , $14^2=196$ , which condracticts the assumption that all the individual digits of $N^2$ are $\leq 6$ . Thus $N=14$ is also rejected. There are no possible values of $N$ in Case 3 .

Case 4: $N^2$ has 4 digits. There are 4 possible factorisations: $1 \times 3333, 3 \times 1111, 11 \times 303$ and $33 \times 101$ . Only the last factorisation results in $N^4$ having 4 digits. Thus $\frac {101-33}{2}=34$ is also a possible value of $N$ $N=34$ .

Case 5: $N^2$ has 5 digits. There is only 1 possible value of N in this case: $N=100$ . It is easy to verify that $100^2+33333=43333$ is not a perfect square. There are no possible values of $N$ in case 5 .

Summing up all 3 possible values of $N$ , we have $1+4+34=39$

Case 1: $N^2<10$ , $N^2+3=M^2$ . $M^2-N^2=(M-N)(M+N)=3$ . So, $M-N=1$ , $M+N=4$ . $N=1,M=2$ is a solution.

Case 2: $10 \leq N^2 \leq 99$ $M^2-N^2=(M-N)(M+N)=33$ . So, $[M-N=3,M+N=11]$ or $[M-N=1,M+N=33]$ . $[N=4,M=7]$ and $[N=16,M=17]$ are solutions, but $16^2>100$ , so it is not a solution.

Case 3: $100 \leq N^2 \leq 999$ , $M^2-N^2=(M-N)(M+N)=333$ . So, $[M-N=3,M+N=111]$ or $[M-N=1,M+N=333]$ or $[M-N=9,M+N=37]$ . $[N=54,M=57]$ , $[N=166,M=167]$ and $[N=14,M=23]$ are solutions, but $166^2,54^2>100$ and $14^2=196$ , so they are not solution.

Case 4: $1000 \leq N^2 \leq 9999$ , $M^2-N^2=(M-N)(M+N)=3333$ . So, $[M-N=3,M+N=1111]$ or $[M-N=1,M+N=3333]$ or $[M-N=33,M+N=101]$ or $[M-N=11,M+N=303]$ . $[N=554,M=557]$ , $[N=1666,M=1667]$ , $[N=34,M=67]$ and $[N=146,M=157]$ are solutions, but $554^2,1666^2,146^2>100$ , so they are not solutions.

Conclusion, the only solution for N is 1,4 and 34. $1+4+34=39$ , done.

Now we know that for $N=1, 2, 3$ the $N^{2}$ has one digit. $N=4, 5, 6, 7, 8, 9$ the number $N^{2}$ has 2 digits. $N=10, 11, ..., 31$, $N^{2}$ has 3 digits and for $N=32, 33, ..., 99$ the number has 4 digits. It is clear that for $N=100$, the resulting number $10000+33333=43333$ is not a perfect square since $208^{2}<43333<209^{2}.$

We show that the last digit of $N$ must be $1, 4, 6, 9.$ Note that the last digit of $N$ is $1, 4, 9, 6, 5, 6, 9, 4, 1, 0$ when the last digit of $N$ is $1, 2, 3,4, 5, 6, 7, 8, 9, 0$ and the last digit of $N^{2}+x$ ($x=3, 33, 333, 3333)$ is $4, 7, 2, 9, 8, 9, 2, 7, 4, 3.$ Hence the last digit of $N$ must be $1, 4, 6, 9$. For $N$ with one digit obviously $N=1$ is the only solution. For $N=4, 6, 9$ $N^{2}+33=49, 69, 114$ so only $N=4$ is good.

For $N^{2}$ with 3 or 4 digits we can assume that $N\in {10, 11, ..., 25, 32,33, ...,81}$ since for $N=26, 27, ..., 31, N^{2}\in [676, 999]$ and for $N=82, 83,..., 99$ $N^{2}\in [6724, 9999].$

For $N\in {10, 11, ..., 25},$ by the assumption abov we only need to consider $N=11, 14, 16, 19, 21, 24$ by the last digit analysis, and we have $N^{2}+333=454, 529=23^{2}, 589, 694, 774, 909.$ But $14^{2}=196,$ contradicting the fact that all digits of $N^{2}$ are at most $6$. Hence no solutions for $N\in {10, 11, ..., 25}.$

For $N\in {32, 33, ..., 81},$ it also suffices to consider those numbers with last digit $1, 4, 6, 9.$ We eliminate $36, 41, 44, 54, 59, 61, 64, 69, 74, 76$ as $N^{2}=1296, 1681, 1936, 2916, 3481, 3721, 4096, 4761, 5476, 5776$ and it can be seen that for each of them not all digits are in the interval $[0, 6].$ Hence only $N=34, 39, 46, 49, 51, 56, 66, 71, 79, 81$ is left and $N^{2}+3333$ is $4489=67^{2}, 4854, 5449, 5734, 5934, 6469, 7689, 8374, 9574, 9894.$ So only $N=34$ is valid.

So the complete set of answers is $N=1, 4, 34$ and the sum is $1+4+34=39.$ Q.E.D.

First consider the units of any number N^2, you find they end in either 1, 4, 5, 6, 9. Because the question says we must add 3 to each digit of N^2 (and carryover is disallowed) we can again consider the units in the square; the numbers above which, when increased by 3, produce another number in the set (otherwise it couldn't be a perfect square) are the unit digit of possible N^2's . Using this, we see any N^2 must end in either 1 (since 1+3 = 4) or 6 (since 6+3 =9)

Now that we've determined that N^2 must end in either 1 or 6 we can deduce the the last digit of the number N must be either 1 or 9 (to get an N^2 ending in 1), or 4 or 6 (to get an N^2 ending in 6).

Now that I had limited my possible solutions I adopted a 'nuts and bolts' method. I try all the squares of numbers within from 1-91, 9-99, 4-94, and 6-96,immediately eliminating any with any digits higher than 6. From there, you can do as the question instructs to each of your remaining possibilities and see which one's work. You'll find that the solutions are * N = 1, 4, and 34 *

Therefore the answer is 39

Trying N=100 does not satisfy the condition above, hence we only need to consider the cases where N is a number from 1 to 99. This also tells us that N^2 cannot be a 5 digit number.

Let N^2 be a 4 digit number and M^2 be the new number. N^2 = 1000A+ 100B+10C+D. Note that M^2 = 1000(A+3)+100(B+3)+10(C+3)+D+3 = 1000A+100B+10C+D+3333. Let's bear in mind that A,B,C and D are integers less than or equal to 6 at the same time.

This gives us M^2-N^2 = (M+N)(M-N) = 3333. Since M+N can only be 3333,1111,303 or 101, M-N can only be 1,3,11 or 33 in this case. Trying all 4 possibilities would only yield the case where M = 67 and N = 34. This gives us 1 solution for N.

Now consider the case when N^2 is a 3 digit number. By the same reasoning in the 2nd paragraph, M^2-N^2 = 333. Using the same method to find M+N and M-N, we realise that there is no solution for N which satisfies the condition as stated in the question.

When N^2 is a 2 digit number, M^2-N^2 = 33. After we find M+N and M-N, we get a solution N=4.

When N^2 is a 1 digit number, M^2-N^2 = 3. this gives us the only solution where N=1.

In total we have 3 solutions, N=1, N=4 and N=34. Therefore the answer is 39.

Let M and N be positive integers with N \leq 100. Since we need to have the difference M^2 - N^2 to comprise only of digits 3 at all possible places, and that N \leq 100 forces N^2 to have at most 4 digits, we need to look at the following four equations:

\begin{align} M^2 - N^2 = 3, \ M^2 - N^2 = 33, \ M^2 - N^2 = 333, \ M^2 - N^2 = 3333. \ \end{align}

I will now write the numbers and expressions in these equations in their factorization in positive integers:

\begin{align} $M + N$ $M - N$ = 3 \times 1, \ $M + N$ $M - N$ = 11 \times 3, \ $M + N$ $M - N$ = 37 \times 3 \times 3, \ $M + N$ $M - N$ = 101 \times 11 \times 3. \ \end{align}

We now split the RHS of each of these equations in two factors, and equate the larger one with $M + N$ and the smaller with $M - N$ .

For the first equation, we have $M + N$ = 3, $M - N$ = 1, which gives M = 2, N = 1. Indeed, 1^2 + 3 = 4 = 2^2.

For the second equation, we have two possibilities: $M + N$ = 11, $M - N$ = 3, giving M = 7, N = 4 $Indeed, 4^2 = 16, 7^2 = 49 = 16 + 33$ ; and $M + N$ = 33, $M - N$ = 1, giving M = 17, N = 16 \( not of use).

Similarly for the third equation, we have three possibilities - although none will work for our purpose: \( M + N \) = 37, $M - N$ = 9 gives M = 23, N = 14. $M + N$ = 111, $M - N$ = 3 gives M = 57, N = 54. $M + N$ = 333, $M - N$ = 1 and N is already greater than 100. $M = 167, N = 166$ .

For the final equation, we have four possibilities: $M + N$ = 101, $M - N$ = 33, yielding M = 67, N = 34, with N^2 = 1156 and M^2 = 4489 = 1156 + 3333. The other three possibilities yield N > 100

Thus, the numbers 1, 4 and 34 satisfy the requirement of our problem, and their sum is 39.

Let $P^2$ = $N^2+3$ for $N^2$ a 1 digit value,

$P^2$ = $N^2+33$ for $N^2$ a 2 digit value,

$P^2$ = $N^2+333$ for $N^2$ a 3 digit value,

$P^2$ = $N^2+3333$ for $N^2$ a 4 digit value,

since $N \leq 100$ and no carryover is allowed so $N^2 \leq 6666$

$P^2 - N^2 = 3$

$(P+N)(P-N) = 1 \times 3$

$P+N = 3$ , $P-N = 1$

$P=2 , N=1$

by using this method we will also found that

$P^2$ = $N^2+33 \rightarrow P=7 , N=4$

$P^2$ = $N^2+333\rightarrow P=57 , N=54$

$P^2$ = $N^2+3333\rightarrow P=67 , N=34$

$P^2$ = $N^2+3333 \rightarrow P=557 , N=554$

since $N=554$ is larger than 100 and

one of the digit of $N^2$ if $N=54$ is exceed 6

therefore $N=554$ and $N=54$ is ignored

the answer will be $34 + 4 + 1 = 39$

N 2 + 3 or 33 or 333, or 333 = M 2

Hence try the following (M+N) * (M-N) = 3X1 => N=1 (M+N) * (M-N) = 33X1 => N=16, but N 2 has 3 digits = 11X3 => N=4 (M+N) (M-N) = 333x1 N > 100 = 111x3 N =54, but N 2 has 4 digits = 9x37 N=14, N 2 = 196 has digit >6 (M+N) (M-N) = 3333 = 33x101 N=34

So suitable Ns are 1,4,34 and their sum = 39

the squares of numbers below 100 are all below 10000, therefore we can add 3,33,333 or3333 to the square of the number. let the numbers be n,o,p,q,r,s,t and u, we have n^2 +3= o^2 p^2 +33= q^2 r^2 +333= s^2 t^2 +3333= u^2 by trial error, n=1 and p=4 can be found easily. for the next use, we apply the Fundamental Theorem of Arithmetic(FTA) we are trying to find numbers a,b,c,d so that ab=333 and cd=3333 since 333= 3^2 x 37 and 3333=3x11x101, the best combination so that the numbers are as low as possible are 9x37 and 33x101. we know that n^2 +2n+1= (n+1)^2 , so by applying this we find out that 18^2 +37 = 19^2 and 50^2 +101= 51^2. we know that (n+2)^2 - (n+1)^2 - (2n+1) = 2, therefore we can change the difference between the squares of numbers into an arithmetic progression where a is the number we wanted and d=2, we will get a+(9-1)/2 d=37 and a+(33-1)/2 d =101, we get a=29 and a=69. therefore, r^2 +29= (r+1)^2 and s^2 +69= (s+1)^2 . we find out that r=14 and s=34. 14 is omitted because the digits on r^2 cannot exceed 6. therefore the sum of all solutions are 34+4+1=39.

From above, we can see that 3, 33, 333, 3333( under 100^2) can be a difference between 2 square numbers, either adjacent or distinct. Let's start with 3, 1=1=1^2 1+3=4=2^2 1+3+5=9=3^2 and so on... so, 1^2+3=2^2 N=1

Then go to 33, 17^2-16^2=33, but the two square numbers have 3 digits, so N\neq 16. 33 can be split into 3 numbers, 33=9+11+13 \frac{9-1}{2}=4, while \frac{13+1}{2}=7 4^2+33=7^2 N=4 Then go to 333, 333 can be split into 109+111+113 \frac {109-1}{2}=54, \frac{111+1}{2}=56 However, 54^2=2916, which contradicts with the question:"There is no carryover allowed. It is an implicit assumption that the individual digits of N^2 are all ≤6." 333 can also be split into 29+31+33+35+37+39+41+43+45 \frac{29-1}{2}=14, \frac{45+1}{2}=23 However, 14^2=196, which contradicts with the question again. Now to 3333, 3333 can be split into 1109+1111+1113 \frac{1109-1}{2}=554, which contradicts with N≤100. 3333 can also be split into 11 numbers, 293+295+...+311+313 \frac{293-1}{2}=146, which contradicts with the question again. Then, 3333 can be split into 33 numbers, 69+71+...+131+133 \frac{69-1}{2}=34, \frac{133+1}{2}=67 34^2+3333=67^2(1156+3333=4489) So N=34. Finally, 34+4+1=39

I basically looked for perfect squares that differed by 3, 33, 333, and 3333. Not all of them will be solutions, but it's a way to find them.

Start with the largest, 3333.

$a^2=b^2+3333$

This becomes

$(a+b)(a-b)=3333$

There are 4 pairs of positive factors of 3333: 1 and 3333, 3 and 1111, 11 and 303, 33 and 101. In each case, the values of a and b are 1667 and 1666, 557 and 554, 157 and 146, 67 and 34 respectively. The last pair is the only one where the numbers are less than 100, so put 34 on our list.

Go through the same procedure with 333, 33, and 3, and you find you will only add 4 and 1 to our list. So our answer is 1+4+34=39.

$N^2 is \leq 10000$ . Increasing each digit of N^2 by the 3 means if $N^2$ is $1,2,3,4,5$ digit number we have to add $3,33,333,3333 ,33333$ and $N \leq 3,9,31,99,100$ respectively . Let the new number be $b^2$ . If $N^2$ is single digit then, $N^2-b^2=3$ or $(b+N)(b-N)=3$ . so $N = 1$ if $N^2$ is double digit: $(b+N)(b-N)=33$ . and $N = 4$ if $N^2$ is triple digit: $(b+N)(b-N)=333$ . divisors of $333=1,3,37,111,333.$ There is no valid solution for this. if $N^2$ is 4 digit: $(b+N)(b-N) = 3333.$ divisors are $1,3,11,101,1111,3333.$ Only valid solution for this is $N=34.$

$43333$ is not a perfect square.

So the sum is $= 39.$

As for n<100 , n^2 can have maximum 4 digit so Case 1 when n n is of 1 digit for n 1 to 3 N N+3=k k So 3=(k-n) (k+n) k-n=1,n+k=3 N=1,k=2 Case 2 when n^2 has 2 digit n is from 4 to 9 So 33=k k-n n 3 11=(k-n) (k+n) K-n=3 K+n=11 K=7,n=4 Case 3 when n n has 3 digit n is from 10 to 31 333=k k-n n also333 should be expressed as 9 37 not 3 111 As k+n is in 2 digit So 9 37=(k-n) (k+n) K-n=9 K+n=37 K=23,n=14 But 14 14=196 we need no digit greater than 6 so it is neglected Case 4 when n n has 4 digit or n is from 32 to99 So 3333=k k-n n 33 101=(k-n)*(k+n) K-n=33 K+n=101 K=67,n=34 So adding all n is 1 , 4 , 34 So ans is 39

example: M^2-N^2=(M-N)(M+N)=3333=33*101 M-N=33 M+N=101 N=34, M=67

We split into 4 cases, according to the number of digits of $N^2$ .

Case 1: $N^2 = \overline{a}$ , with $a \leq 6$ . Let $M^2 = N^2 + 3$ , and we have that $M <\sqrt{10} < 3.17$ . Then, $(M-N)(M+N) = 3$ which means that $M+N=3, M-N=1$ , hence $M=2, N=1$ is the only solution. We check that $N^2=1$ satisfies the condition that $a\leq 6$ .

Case 2: $N^2 = \overline{ab}$ , with $a, b \leq 6$ . Let $M^2 = N^2 + 33$ , and we have that $M < \sqrt{100} = 10$ . Then, $(M-N)(M+N) = 33$ . Since the sum of the 2 factors equals $2M$ which is less than 20, the only possibility is $M+N=11, M-N=3$ , giving $M=7, N=4$ . We check that $N^2 = 16$ satisfies the assumption that $a, b \leq 6$ .

Case 3: $N^2 = \overline{abc}$ , with $a, b, c \leq 6$ . Let $M^2=N^2+333$ and we have that $M< \sqrt{1000} < 31.7$ . Then $(M-N)(M+N)=3\cdot 3\cdot 37$ . Since the sum of the 2 factors equals $2M$ which is less than $64$ , the only possibility is $M-N=9$ and $M+N=37$ , giving $M=23$ and $N=14$ . However $14^2=196$ , violating the assumption that $b \leq 6$ .

Case 4: $N^2 = \overline{abcd}$ , with $a, b, c, d \leq 6$ . Let $M^2=N^2+3333$ and we have that $M < \sqrt{10000}=100$ . Then $(M-N)(M+N)=3333=3\cdot 11\cdot 101$ . Since the sum of the 2 factors equals $2M$ which is less than $200$ , the only possibility is $M-N=33$ and $M+N=101$ , giving $M=67$ and $N=34$ $\Rightarrow N^2 =34^2=1156$ . This satisfies the condition that $a, b, c, d \leq 6$ , hence $N=34$ .

Therefore, the numbers $N= 1, 4, 34$ satisfies the given conditions and the sum is equal to $39$ .

(Note: Sorry, I pretty much rushed in making this solution but I guess it is evident.)

Let a= b \cdot c where b and c are positive factors of a and b > c Let N^2+a=k^2 for some positive integer k and N \rightarrow k^2-N^2=a \rightarrow (k+N)(k-N)=b \cdot c Since (k+N) > (k-N) we equate k+N=b and k-N=c solving for N, we get N= \frac {b-c}{2}

We then take only the possible cases of (b,c): Case 1: 1 \leq N < \sqrt{10} < 4, a=3 We get (b,c)=(3,1) We get N=1 Case 2: 4 \leq N < \sqrt{100}=10, a=33 We get (b,c)=(11,3) We get N=4 Case 3: 10 \leq N < \sqrt{1000} < 32, a=333 We get (b,c)= (111,3), (37,9) We get N=54,14 Case 4: 32 \leq N < \sqrt{10000} = 100, a=3333 We get (b,c)= (101,33) We get N=34 Case 5: 100 \leq N < 100 < \sqrt{100000}, a=33333 We get no possible pair (b,c) that yields a possible N (We can get N=74 but 74>100 thus having a contradiction in the assumption) After checking, since 54^2 and 14^2 have digits higher that 6, they are ineligible since it would cause the digits to carry over.

Therefore, the only possible N are 1,4, and 34. The sum of which is 39.