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Question

Which of the integers 10, 11, 12 and 13 can be written as the sum of squares of four integers (allowing repetition)?

The correct answer is

All

Integers as Sum of Four Squares

The question asks which of the integers 10, 11, 12, and 13 can be written as the sum of the squares of four integers, allowing repetition. This is a classic problem in number theory related to representing numbers as sums of squares.

According to Lagrange's four-square theorem, every natural number can be represented as the sum of four integer squares. This means that all integers, including 10, 11, 12, and 13, should be expressible in this form.

Let's demonstrate how each of these integers can be written as the sum of four squares:

10 as Sum of Four Squares

We need to find four integers a, b, c, and d such that $a^2 + b^2 + c^2 + d^2 = 10$. We can use squares of small integers like $0^2=0$, $1^2=1$, $2^2=4$, $3^2=9$, etc.

One way to express 10 as the sum of four squares is:

  • $\text{10} = 3^2 + 1^2 + 0^2 + 0^2 = 9 + 1 + 0 + 0$

Since we found a combination, 10 can be written as the sum of four squares.

11 as Sum of Four Squares

Next, we check if 11 can be written as the sum of squares of four integers. Using squares 0, 1, 4, 9:

We look for a combination $a^2 + b^2 + c^2 + d^2 = 11$. Here is one possibility:

  • $\text{11} = 3^2 + 1^2 + 1^2 + 0^2 = 9 + 1 + 1 + 0$

Thus, 11 can also be written as the sum of four squares.

12 as Sum of Four Squares

Now, let's examine the integer 12. Can it be written as the sum of squares of four integers?

We need $a^2 + b^2 + c^2 + d^2 = 12$. Here is one way:

  • $\text{12} = 2^2 + 2^2 + 2^2 + 0^2 = 4 + 4 + 4 + 0$

Another possible way is:

  • $\text{12} = 3^2 + 1^2 + 1^2 + 1^2 = 9 + 1 + 1 + 1$

Since we found ways to express 12 in this form, 12 can be written as the sum of four squares.

13 as Sum of Four Squares

Finally, let's check the integer 13. Can 13 be written as the sum of squares of four integers?

We need $a^2 + b^2 + c^2 + d^2 = 13$. Consider this combination:

  • $\text{13} = 3^2 + 2^2 + 0^2 + 0^2 = 9 + 4 + 0 + 0$

This combination works, confirming that 13 can be written as the sum of four squares.

Sum of Squares Conclusion

We have shown examples for each integer:

  • 10 = $3^2 + 1^2 + 0^2 + 0^2$
  • 11 = $3^2 + 1^2 + 1^2 + 0^2$
  • 12 = $2^2 + 2^2 + 2^2 + 0^2$
  • 13 = $3^2 + 2^2 + 0^2 + 0^2$

Therefore, all the integers 10, 11, 12, and 13 can be written as the sum of squares of four integers.

The option that states 'All' is correct because all the given integers can be represented in this way.

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Important Questions from Integers

  1. The average of eleven consecutive positive integers is d. If the last two numbers are excluded, by how much will the average increase or decrease?
  2. The numerator of fraction is 3 more than the denominator. When 5 is added to the numerator and 2 is subtracted from the denominator, the fraction becomes 8/3, When the original fraction is divided by \(5 \frac{1}{2}\) , the fraction so obtained is:

  3. The sum of a non - zero number and twenty times its reciprocal is 9. What is the number?

  4. If \(\frac{{45}}{{53}} = \frac{1}{{a + \frac{1}{{b + \frac{1}{{c - \frac{2}{5}}}}}}},\)  where a, b and c are positive integers, then what is the value of (4a - b + 3c)

  5. The denominator of a fraction is 4 more than the double of its numerator. When 3 is added to the numerator and 3 is subtracted from denominator the fraction becomes 2/3. Then find the difference between denominator and numerator of the original fration. 

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