Prove that $\frac{21n-3}{4}$ and $\frac{15n+2}{4}$ cannot both be integers for the same positive integer $n$.
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3Add the two numbers. What do you get? – Andrés E. Caicedo Apr 08 '14 at 00:29
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@AndresCaicedo: that works out nicely so that you don't need to double as in my answer. – robjohn Apr 08 '14 at 00:38
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Hint $\ $ If $\,4\,$ divides $\,21n-3\,$ and $\,15n+2\,$ then it divides their sum $\,36n-1,\,$ contradiction.
Remark $\ $ In parity language: $ $ integers divisible by $\,4\,$ are even so their sum is even. However, their sum $\,=\, 36n-1\,$ is $ $ odd, $ $ a contradicton. Or, equivalently, their difference $\,= 6n-5\,$ is odd (see robjohn's answer for a fractional version of this).
Bill Dubuque
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1A bit cleaner than my answer since the doubling is not necessary. (+1) However, is yours really a hint? :-) – robjohn Apr 08 '14 at 00:39
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If they are both integers, then so is their difference which is equal to $\frac{6n-5}4$, absurd since the numerator is odd.
Sandeep Silwal
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In mod 4, $$21n-3=n+1$$ and $$15n+2=3n+2$$ If $n+1=0$ then we would have $n=-1$ and hence $$3n+2=-1$$ so the second quantity could not be divisible by 4. Conversely, $3n+2=0$ would imply $n=2$ so that $$n+1=3$$ would not be divisible by 4. Thus they cannot both be integers.
user140943
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The sum of two integers is an integer.
The sum of these two numbers $$\frac{21n - 3}{4} + \frac{15n + 2}{4} = \frac{36n - 1}{4} = 9n-\tfrac{1}{4}$$ is obviously not an integer.
Since the sum is not an integer, the addends can't be, either.
SQB
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