# Prove that #if u# is an odd integer, then the equation #x^2+x-u=0# has no solution that is an integer?

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This question comes from a section on contradiction and proof by cases. I thought that I could use the logical equivalency #p=>q-=notq=>notp# , but I'm sure how to go about showing that something *has* an integer solution either in the case of this particular equation.

Any guidance would be much appreciated.

(This question is from Mathematical Reasoning by Ted Sundstrom, Sect. 3.4 Q 2)

This question comes from a section on contradiction and proof by cases. I thought that I could use the logical equivalency *has* an integer solution either in the case of this particular equation.

Any guidance would be much appreciated.

(This question is from Mathematical Reasoning by Ted Sundstrom, Sect. 3.4 Q 2)

##### 3 Answers

#### Answer:

Hint 1: Suppose that he equation

#### Explanation:

If

Where

But the second equation entails that

Now, both

**Proposition**

If

**Proof**

Suppose that there exists a integer solution

#x^2 + x - u = 0#

where

#m# is odd; or

#m# is even.

First, let us consider the case where

# m = 2k + 1 #

Now, since

# m^2 + m - u = 0 #

# :. (2k + 1)^2 + (2k + 1) − u = 0 #

# :. (4k^2 + 4k + 1) + (2k + 1) − u = 0 #

# :. 4k^2 + 6k + 2 − u = 0 #

# :. u = 4k^2 + 6k + 2 #

# :. u = 2(2k^2 + 3k + 1) #

And we have a contradiction, as

Next, let us consider the case where

# m = 2k #

Similarly, since

# m^2 + m - u = 0 #

# :. (2k)^2 + (2k) − u = 0 #

# :. 4k^2 + 2k − u = 0 #

# :. u = 4k^2 + 2k #

# :. u = 2(2k^2 + k) #

And, again, we have a contradiction, as

So we have proved that there is no integer solution of the equation

Hence the proposition is proved. QED

#### Answer:

See below.

#### Explanation:

If