Finding a Perfect Square

Let the given number $N = m^2$ and let the number formed increasing the first digits of N be $q^2$ . We are given that $N$ is increased by 123 to obtain a new square i.e $q^2 - m^2 = 123$ , $(q + m) (q- m) = 123$ .

Here we observe that 123 can be written only as $123 *1$ or $41*3$ as positive factors. Since $q+m > q - m$ , this implies $(q+ m) = 123$ or 41 and $(q-m)= 1$ or 3 respectively. On equating this linear equation, We get $m = 19$ or 62. As $m^2$ is a three digit number, we reject $m = 62$ , therefore $m = 19 , N = m^2 = 361$ .

[Latex edits - Calvin]

$N=x^2 ; x^2+123=y^2 \\ y^2 - x^2 =123 \\ (y+x)(y-x)=41\times 3 => y+x=41 ; y-x=3 \\ x = 19 => N=361$

Notice that among $1^2, 2^2, \cdots , 10^2$ , the units digits are $1,4,5,6,9,0$ . Therefore, these are the only quadratic residues mod $10$ . The only pairs of these that are $3$ apart are $1,4$ and $6,9$ .

The three digit perfect squares range from $11^2 = 121$ to $31^2 = 961$ . Therefore, the only possible pairs of squares that could possibly work for $N$ are the ones ending in $1$ or $6$ , which are $11^2, 14^2, 16^2, 19^2, 21^2, 24^2, 26^2, 29^2, 31^2$ . It is now easy to verify that $361$ is our answer.

Since we know that the digit increases do not regroup, we know that the resulting number is N+123, which is also a perfect square. Letting N = k k, we know that k k+123 = m*m for some m > 0 and integer. Then m^2-k^2 = 123, which means that (m+k)(m-k) = 123. Since m+k and m-k are positive integers, either m+k = 41 and m-k = 3, or m+k = 123 and m-k = 1. It is easy to see that the second case gives no solution, and the first case gives k = 19 and m = 22. Thus N = 361, and if we perform the procedure on 361, we get 484, another square.

By increasing the first digit by 1, second digit by 2, third digit by 3 and using the assumption given, N increases by 123. Let $N=a^2, N+123=b^2$ for some $a,b \in \mathbb{Z}^+$ Therefore, $b^2-a^2 = 123 \Rightarrow (b+a)(b-a)=3*41$ . Clearly $b+a > b-a > 0$ and the factors are integers so there are only 2 cases.

Case 1: $b+a = 123, b-a = 1$ Solving we get $a=61,b=62$ which is not possible since $N=a^2=61^2>1000$ so it is not a 3-digit number.

Case 2: $b+a = 41, b-a = 3$ Solving we get $a=19,b=22$ so $N=19^2=361$ and $N+123 = 22^2 = 484$ so the required answer is 361.

Let $N=a^2, X=b^2, a,b$ are natural numbers, where $X$ is the larger perfect square. Note that $b^2-a^2=(b+a)(b-a)=123=41 \times 3=1 \times 123$ . So there are 2 possible pairs of linear equations to solve. $(a,b)=(22,19), (62,61)$ . Clearly we are finding the former pair, since for the latter pair $x=61^2$ is not 3 digit. So $N=19^2=361$ .

First we take all the 3 digit numbers under 900 which are perfect square. we find 21 numbers, they are 100,121,144,169,196,225,256,289,324,361,400,441,484,529,576 ,625,676,729,784,841,900. after checking we get 361 only number in which if we add ,1to its hundred place ,2 to its tens place, 3 to its ones place then the result is also a perfect square. 361+123=484 which is a perfect square. hence 361 is the answer.

Let the new number after given operations be M

Obviously, M-N=123

Case1: M=N+1

${(N+1)}^2$ - ${N}^2$ =123

⇒N=62

As, ${N}^2$ >999

So,62 is rejected

Case2: M=N+2

${(N+2)}^2$ - ${N}^2$ =123

⇒N is fraction

So,Case2 is rejected

Case3: M=N+3

${(N+3)}^2$ - ${N}^2$ =123

⇒N=19

${N}^2$ =361

Hence, _ Ans.=361 _

suppose that N = y^2, and the new perfect square is x^2, then x^2-y^2 = 123 <=> (xy) (x + y) = 3 * 41. with elimination we obtained that x = 22 and y = 19. so N = 19^2 = 361

Let $N=x^2$ . Then $x^2+123=y^2$ for some positive integer $y$ as given.

Subtracting and factoring, we have $(y+x)(y-x)=123=3*41$ , and since both $x$ and $y$ are positive $y+x=41$ and $y-x=3$ . Solving, we get $x=19$ , which means $N=x^2=361$ .

Let's say $N=a^2$ , and $N+123=B=b^2$ . Then, $B-N=123$ $\Rightarrow$ $(b+a) \cdot (b-a) = 123 = 3 \cdot 41$ . Notice that $(b+a) < 3 \cdot 41$ and $(b-a) > 1$ , so we have, by the Fundamental Theorem of Arithmetic: $b+a = 43$ and $b-a = 3$ . Then, we conclude that $a=19$ , from what one gets $N=a^2=361$ .

abc = N^2

(a + 1)(b + 2)(c + 3) = k^2

100a + 10b + c = N^2 -----------(1)

100a + 100 + 10b + 20 + c + 3 = k^2--------(2)

put (1) in (2)

N^2 - k^2 = -123

(N - k)(N + k) = -3 * 41 or -41 * 3

according to question its clear that k is a bigger number

so , N - k = -3 ---------------(3)

N + k = 41 --------------(4)

add both (3) and (4)

N = 19 and k = 22

required answer = 19^2 = 361

Let that number be \overline{xyz} .... So converting it into digit order it can be written as 100x + 10y + z = N As this number is a perfect square, Let N = k^2 \Rightarrow Now we will increase the digits as given

First digit is increased by 1,so it will become x + 1, Second digit is increased by 2,so it will be y + 2, Similarly third digit will become z + 3, Finally our new number is \overline{x+1 y+2 z+3}

Now this new number is also a perfect square, so let it be p^2.

writing in digit order,

100(x + 1) + 10(y + 2) + z +3 = p^2

100x + 100 + 10y + 20 +z + 3 = p^2

Now 100x + 10y + z can be replaced by k^2

k^2 + 123 = p^2

p^2 - k^2 = 123

(p+k)(p-k) = 123 We know that p> k, so p+k > p-k now 123 can be written as 41 \times 3, So p+k = 41 p-k= 3 On solving we get p=22, k=19 So N = k^2 = 361.

Let $N = p^2$ Then, $N+123 = q^2$

where p and q are positive integers.

$q^2 - p^2 = 123 = 3*41$ $(q-p)*(q+p) = 3*41$

Since, p and q are both positive integers hence p = 19 & q = 22 $N = p^2 = 361$

N = c + 10b + 100a = X² M = (c+3)+10(b+2)+100(a+1)=Y² M=N+123=Y² M = X²+123=Y² 123 = Y²-X² 123=(Y-X)(Y+X) Y-X = 1 and Y+X = 123 .. X = 61 .. X²=N=3721 Not a 3-digit number Y-X=3 and Y+X=41 ... X=19 .. X²=N=361

"n is a 3-digit number" Let "abc" represent the digits of n. 0 < a n = 100a + 10b + c

"n is a perfect square." n = 100a + 10b + c = p², where p is an integer.

"When the first digit is increased by 1, the second digit is increased by 2, and the third digit is increased by 3, the result is still a perfect square." 100(a+1) + 10(b+2) + (c+3) = q²

expand terms 100a + 100 + 10b + 20 + c + 3 = q²

regroup terms (100a + 10b + c ) + 123 = q² p² + 123 = q² q² - p² = 123 (q + p)(q - p) = 123 123 is the product of two prime numbers, 3 and 41, and q > p, so q + p = 41 q - p = 3

q = 22 q² = 484 n = p² = q² - 123 = 361

N may be written as 100a + 10b + c, where a, b, and c are the digits of N. The altered number is equal to $100(a + 1) + 10(b + 2) + (c + 3) = 100a + 10b + c + 123 = N + 123$ . Since $N = k^{2}$ and the altered number is equal to $m^{2}$ for positive integers k and m, $k^{2} + 123 = m^{2}$ . Rearranging and factoring, $(m + k)(m - k) = 123$ . Since the factorization of 123 is 3 * 41, either m + k = 123 and m - k = 1 or m + k = 41 and m - k = 3. In the first case, m = 62 and k = 61, but $61^{2}$ is not a 3-digit number. Thus the second case must be the correct solution (m = 22 and k = 19), so $N = 19^{2} = 361$ .

Let N be $n^{2}$ for some integer n.

$n^{2}$ + 123 = $k^{2}$ for some integer k that is strictly greater than n.

Rearranging the equation,

$k^{2}$ - $n^{2}$ = 123

Factorising the LHS and RHS,

(k - n)(k + n) = 3 x 41

Evidently, k - n = 3 and k + n = 41 since k > n.

Solving the equations, one will get n = 19

Thus, N = 19 x 19 = 361

Perfect squares differ from each other by consecutive odd numbers. Since 123 has a prime factorization of 41 x 3, it can be written as 39 + 41 + 43. This means that to find N, we need to find the perfect square that differs from the next one above it by 39. 400 - 361 = 39, so N = 361. Indeed, 361 is 19 squared and 361 + 123 = 484 is 22 squared.

Since 3-digits perfect square is only about 22 numbers, so trial and error can be used in this problem.

$1. N$ should be less than 800

$2.$ The last digit of perfect square is 1,4,5,6, and 9.

To make the last digit of $N$ + 3 is also perfect square, the last digit of $N$ should be 1.

Let us list the possible numbers : 121; 361; 441; 841

By trial and error method :

121+123 = 244

361+123=484

441+123=564

841+123=964

From the result we can see that the correct answer for $N$ is 361 as 484 = $22^2$

N é um número de 3 algarismos, então N pode ser escrito na forma: "XYZ" (sendo X, Y e Z os algarismos que o compõem). Neste caso, seu valor seria N = 100X + 10Y + Z. N é um quadrado perfeito, N = a². Então: a² = 100X + 10Y + Z .

"Quando o primeiro dígito é aumentado por 1, o segundo dígito é aumentado por 2, o terceiro dígito é aumentado por 3, o resultado é ainda um quadrado perfeito."

100(X + 1) + 10(Y + 2) + Z + 3 = b² (b², pois é um outro quadrado perfeito).

b² = 100(X + 1) + 10(Y + 2) + Z + 3

b² = 100X + 100 + 10Y + 20 + Z + 3

b² = 100X + 10Y + Z +123 ⇒ b² = a² + 123

b² - a² = 123

(b - a)(b + a) = 123

Observação: O número 123 = $41 \times 3$ , em que ambos são primos.

Como b > a, então temos que:

b + a = 41

b - a = 3

Então b = 22 e a = 19.

Conclusão: N = a² = 19² = 361.

as the perfect square end with 1,4,6,9 by adding 3 to the unis place only we again get perfect square so by checking all these we get answer

O quadrado deve estar entre 100 e 876, pois o maior algarismo que compõe o número é 9. Portanto começamos por 10² = 100, fazendo as alterações que o exercício manda, chagamos em 223, que não é quadrado perfeito. Então segue para 11² = 121, se alterarmos, temos 244, que também não é quadrado perfeito, e fazemos até chegar a 19² = 361, cujo alteração nos levará a 484 = 22²

assume N = a^2 and N + 123 = b^2 a^2 + 123 = b^2 b^2-a^2 = 123 (b+a)(b-a)= 123 , 123 = 1 X 123 ,
123 = 3 X 41 Because b^2-a^2 = 123, and a,b≥0 , we get b>a

- Assume (b+a)=123 and (b-a)=1

We get b=62 and a=61, so N = 61^2 = 3721 Contradiction.

- Assume (b+a)=41 and (b-a)=3

We get b=22 and a=19, so N = 19^2 = 361 So N = 361

paragraph 1 Let N = 100a + 10b + c = x^{2} . After the digits of N are changed (in the question), N = 100(a+1) + 10(b+2) + c+3 = 100a + 10b +c + 123 = x^{2} +123 = y^{2} . paragraph 2 Since, y > x, y^{2} = (x + k)^{2} = x^{2} + 2kx + k^{2} for some positive integer k. Note that 2kx + k^{2} = 123. Factoring k out, we get k(2x +k) = 123. Using prime factorisation of 123, we get k(2x +k) = 123 = 3 \times 41 , since both k and 2x+k are clearly integer. k < 2x+k , therefore, we get k = 3 and x = \frac{41 - 3}{2} = 19. N = 19^{2} = 361. Done. :)

"n is a 3-digit number" Let "abc" represent the digits of n. 0 < a n = 100a + 10b + c

"n is a perfect square." n = 100a + 10b + c = p², where p is an integer.

"When the first digit is increased by 1, the second digit is increased by 2, and the third digit is increased by 3, the result is still a perfect square." 100(a+1) + 10(b+2) + (c+3) = q²

expand terms 100a + 100 + 10b + 20 + c + 3 = q²

regroup terms (100a + 10b + c ) + 123 = q² p² + 123 = q² q² - p² = 123 (q + p)(q - p) = 123 123 is the product of two prime numbers, 3 and 41, and q > p, so q + p = 41 q - p = 3

q = 22 q² = 484 n = p² = q² - 123 = 361

Let a,b,c be the digits. abc = x^2. abc + 123 = y^2. Substracting first one form second one, 123 = y^2-x^2 = (y+x)(y-x) = 41* 3. Hence y+x=41, y-x = 3. Solving, y = 22. Therefore, abc + 123 = 484. So, abc = 361

Let $N= n^2$ . So, $n^2+123=m^2$ . By algebra we have $123=(m-n)(m+n)$ where $m-n$ and $m+n$ are integers. The prime factorization of $123$ is $3*41$ . Therefore, $m-n=1$ and $m+n=123$ or $m-n=3$ and $m+n=41$ .

If $m-n = 1$ and $m+n = 123$ , then we have $m=62$ and $n=61$ . But then, $N=n^2=3721$ , which contradicts that $N$ is a 3-digit number.

We then check the other possibility: $m-n=3$ and $m+n=41$ . We get $m=22$ and $n=19$ . Verifiying our answer we see $19^2+13=484=22^2$ . Thus, $N=n^2=361$ .

N^2 + 123 = M^2.
M^2 - N^2 = 123.
M^2 - N^2 = 39 + 41 +43 (Since every two consecutive perfect squares X and Y have a difference of (2 * root X) - 1).
21^2 - 19^2 = 123.
so, N = 19.
and N^2 = 361.
Hence, the answer is 361