# The sum of three natural m=numbers is 2401. The sum of every two of the three numbers is a perfect square. Find the numbers.

1
by mugdha9801

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by mugdha9801

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Perhaps there are multiple solutions. The solution i have is 720, 880 and 801.

720 + 880 + 801 = 2401, 720 + 880 = 40², 880 + 801 = 41²

and 720 + 801 = 39²

=====================================================

Let the three natural numbers be M , N , P. Let there be some natural numbers x, y, and z such that the following is true.

M + N + P = 2401 = 49² -- equation 1

M + N = x² -- eq 2

N + P = y² -- eq 3

P + M = z² -- eq 4

Since the above sums are less than 2401,__x, y, and z are less than 50 __and around 40.

Sum these above three equations and/ or subtract to get the following :

**x² + y² + z² = 2 * (M+N+P) = 4802 -- eq 5**

x² + y² - z² = 2 N --- equations 6

y² + z² - x² = 2 P

z² + x² - y² = 2 M -- eq 8

x² + y² = 2401 + N = 49² + N -- eq 9

y² + z² = 2401 + P = 49² + P

z² + x² = 2401 + M = 49² + M -- eq 11

Hence, we get

N = 49² - z² = (49 - z) (49 + z) -- equation 12

P = 49² - x² = (49 - x)(49 + x) -- eq 13

M = 49² - y² = (49 - y) (49 + y) -- eq 14

From the above, we know that , M, P and N are not prime. They have factors.

Further, we get the following equations 15 to eq 17:

(x y z)² = (49² - M) (49² M + NP) = (49² - N)(49² N + MP)

= (49² - P) (49² P + MN)

We get equations 18 to 20 as

y² z² = 49² P + M N , x² y² = 49² N + M P,

x² z² = 49² M + P N

==========================================

Now we solve the given problem as below. From equation 5, we get that x, y and z are all even or two of them odd and one of them is even. Then only the sum on RHS will be even. On RHS, we have 4802, which is not multiple of 4. Hence, x, y, z are not all even. Two of them are odd. One is even.

M, N, P are not equal among themselves, as per equations 2, 3, and 4, we get that if they are equal 2M = x^2. it is not possible. =========================================

Let us take*x² + y² + z² = 4802* -- equation 5

Let us look at the units digits of x, y and z that can make units digit 2 on RHS.

A square number has units digit as 0, 1, 4, 5, 6, 9 only. Three possible units digits of x², y², and z² which will give 2 as units digit in the sum are :

( 4, 9, 9 ), (0, 1, 1), (1, 5, 6) and (0, 6, 6)

Here (0, 6, 6) have all even. So we dont consider that.

The possible units digits of x, y, and z are the following. As the problem is symmetric about, M, N and P and also about x, y, and z, we only look at the combinations of values.

__Possible units digits a, b, c of x, y, z respectively are listed as:__

(2, 3, 3), (2, 7, 3), (2, 7, 7), (8, 3, 3), (8, 7, 3), (8, 7, 7), (0, 1, 1),

(0, 1, 9), (0, 9, 9),(1, 5, 6), (1, 5, 4), (9, 5, 4), (9, 5, 6)

The equation 5, can be rewritten as

(10 x' + a)² + (10 y' + b)² + (10 z' + c)² = 4802

Here__x', y', and z' are between 0 and 4 only__. as x, y, z are < 50.

100 (x'² +y'² + z'²) + 20 ( a x' + b y' + c z') = 4802 - (a²+b²+c²)

5 (x'² + y'² + z'²) + (a x' + b y' + c z') = 240 + [ 2 - a² - b² - c² ] / 20 -- equation 23

__Hence a² + b² + c² - 2 must be a multiple of 20.__

We check the above combinations and see what matches this criteria.

I find that (0, 1, 9) matches. a =0, b = 1, c = 9.

then 5 (x'² + y'² + z'² ) + ( y' + 9 z' ) = 244

so y' + 9 z' = 4 (mod 5) ie., 4, 9, 14, 19, 24, 29 etc.

I choose, x' = 4 y' = 4, z = 3 , that works.

So x = 10 x' + a = 40,

y = 10 y' + b = 41 and

z = 10 z' + c = 39

*Hence, M = 720, N = 880, P = 801 using equations 12, 13 and 14*.

720 + 880 + 801 = 2401, 720 + 880 = 40², 880 + 801 = 41²

and 720 + 801 = 39²

=====================================================

Let the three natural numbers be M , N , P. Let there be some natural numbers x, y, and z such that the following is true.

M + N + P = 2401 = 49² -- equation 1

M + N = x² -- eq 2

N + P = y² -- eq 3

P + M = z² -- eq 4

Since the above sums are less than 2401,

Sum these above three equations and/ or subtract to get the following :

x² + y² - z² = 2 N --- equations 6

y² + z² - x² = 2 P

z² + x² - y² = 2 M -- eq 8

x² + y² = 2401 + N = 49² + N -- eq 9

y² + z² = 2401 + P = 49² + P

z² + x² = 2401 + M = 49² + M -- eq 11

Hence, we get

N = 49² - z² = (49 - z) (49 + z) -- equation 12

P = 49² - x² = (49 - x)(49 + x) -- eq 13

M = 49² - y² = (49 - y) (49 + y) -- eq 14

From the above, we know that , M, P and N are not prime. They have factors.

Further, we get the following equations 15 to eq 17:

(x y z)² = (49² - M) (49² M + NP) = (49² - N)(49² N + MP)

= (49² - P) (49² P + MN)

We get equations 18 to 20 as

y² z² = 49² P + M N , x² y² = 49² N + M P,

x² z² = 49² M + P N

==========================================

Now we solve the given problem as below. From equation 5, we get that x, y and z are all even or two of them odd and one of them is even. Then only the sum on RHS will be even. On RHS, we have 4802, which is not multiple of 4. Hence, x, y, z are not all even. Two of them are odd. One is even.

M, N, P are not equal among themselves, as per equations 2, 3, and 4, we get that if they are equal 2M = x^2. it is not possible. =========================================

Let us take

Let us look at the units digits of x, y and z that can make units digit 2 on RHS.

A square number has units digit as 0, 1, 4, 5, 6, 9 only. Three possible units digits of x², y², and z² which will give 2 as units digit in the sum are :

( 4, 9, 9 ), (0, 1, 1), (1, 5, 6) and (0, 6, 6)

Here (0, 6, 6) have all even. So we dont consider that.

The possible units digits of x, y, and z are the following. As the problem is symmetric about, M, N and P and also about x, y, and z, we only look at the combinations of values.

(2, 3, 3), (2, 7, 3), (2, 7, 7), (8, 3, 3), (8, 7, 3), (8, 7, 7), (0, 1, 1),

(0, 1, 9), (0, 9, 9),(1, 5, 6), (1, 5, 4), (9, 5, 4), (9, 5, 6)

The equation 5, can be rewritten as

(10 x' + a)² + (10 y' + b)² + (10 z' + c)² = 4802

Here

100 (x'² +y'² + z'²) + 20 ( a x' + b y' + c z') = 4802 - (a²+b²+c²)

5 (x'² + y'² + z'²) + (a x' + b y' + c z') = 240 + [ 2 - a² - b² - c² ] / 20 -- equation 23

We check the above combinations and see what matches this criteria.

I find that (0, 1, 9) matches. a =0, b = 1, c = 9.

then 5 (x'² + y'² + z'² ) + ( y' + 9 z' ) = 244

so y' + 9 z' = 4 (mod 5) ie., 4, 9, 14, 19, 24, 29 etc.

I choose, x' = 4 y' = 4, z = 3 , that works.

So x = 10 x' + a = 40,

y = 10 y' + b = 41 and

z = 10 z' + c = 39