Prove that $n^2 + 3n + 5$ is never divisible by 121 for any positive integer ${n}$.
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Let us say there were indeed a whole number k such that n² + 3n + 5 = 121k. Then n² + 3n + (5 - 121k) = 0.
Then by the quadratic formula, n = [-3 +/- √(9 + 484k - 20)] / 2.
So the proof of the proposition amounts to proving that the discriminant is never a perfect square. That is, 484k - 11 is never a perfect square.
Note that of course, 484 = 22², i.e. 22²k - 11 is never a perfect square.
I think this is clear because if it were, then it would have to be the square of an odd multiple of 11, which means it would have to be a product of 121 and another odd number.
Which means that (484k - 11)/121 would have to be a whole (odd) number. But this is impossible because this is 44k - 1/11, where 44k is whole, therefore the difference is not. Thus we obtain a contradiction, by which we have shown that the discriminant cannot be a perfect square, therefore there is no whole-number solution of n satisfying the initial proposition.
There's an alternate method too:
The given function: f(n) = n^2 + 3n + 5
If divisible by 121 (assume) then:
f(121) = (121)^2+3(121) + 5 should be equal to 0 but no, its equals to 15009
hence, its is not divisible by 121
AND also, RHS =121 is a positive value and LHS is also a positive value, hence another reason it can never be divisible by 121
Or,
f(n) = n² + 3n + 5
f(121) = 121² + 3(121) + 5
f(121) = 14641 + 363 + 5
f(121) = 15009
f(n) is divisible by n if and only if f(n) mod n = 0. Since 15009 is not divisible by 121, f(n) is never divisible by 121
Hope This Helps :)
Then by the quadratic formula, n = [-3 +/- √(9 + 484k - 20)] / 2.
So the proof of the proposition amounts to proving that the discriminant is never a perfect square. That is, 484k - 11 is never a perfect square.
Note that of course, 484 = 22², i.e. 22²k - 11 is never a perfect square.
I think this is clear because if it were, then it would have to be the square of an odd multiple of 11, which means it would have to be a product of 121 and another odd number.
Which means that (484k - 11)/121 would have to be a whole (odd) number. But this is impossible because this is 44k - 1/11, where 44k is whole, therefore the difference is not. Thus we obtain a contradiction, by which we have shown that the discriminant cannot be a perfect square, therefore there is no whole-number solution of n satisfying the initial proposition.
There's an alternate method too:
The given function: f(n) = n^2 + 3n + 5
If divisible by 121 (assume) then:
f(121) = (121)^2+3(121) + 5 should be equal to 0 but no, its equals to 15009
hence, its is not divisible by 121
AND also, RHS =121 is a positive value and LHS is also a positive value, hence another reason it can never be divisible by 121
Or,
f(n) = n² + 3n + 5
f(121) = 121² + 3(121) + 5
f(121) = 14641 + 363 + 5
f(121) = 15009
f(n) is divisible by n if and only if f(n) mod n = 0. Since 15009 is not divisible by 121, f(n) is never divisible by 121
Hope This Helps :)
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