Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $ab = 2$ and $a+b = 5$ then calculate the value of $a^4+b^4$

My approach: $$a^4+b^4 = (a+b)^4-4a^3b-6a^2b^2-4ab^3$$ $$=(5)^4-6(ab)^2-4ab.a^2-4ab.b^2$$ $$=(5)^4-6(24)-4ab(a^2-b^2)$$ $$=(5)^4-6(24)-8(a+b)(a-b)$$ $$=(5)^4-6(24)-8(5)(a-b)$$ I am a little stuck now and any help will be appreciated.

share|cite|improve this question
Check your factorisation $-4ab\cdot a^2-4ab \cdot b^2 = -4ab(a^2+b^2)$. – Macavity May 2 '14 at 10:59
Another possible way of solving this could be using the following factorization: $$a^4+b^4=(a+b)(a^3+b^3)-ab(a^2+b^2)$$ – user26486 May 2 '14 at 11:35
up vote 3 down vote accepted

\begin{align} a^4+b^4 &= (a+b)^4-4a^3b-6a^2b^2-4ab^3 \\&= (a+b)^{4} - 4 ab (a^{2} + b^{2}) - 6 (ab)^{2} \\&= (a+b)^{4} - 4 ab ((a+b)^{2} - 2 ab) - 6 (ab)^{2} \\&= (a+b)^{4} - 4 ab (a+b)^{2} + 2 (ab)^ {2}. \end{align}

As noted in a comment, you made a sign error (it happens) in your calculations, otherwise you would have got here yourself.

share|cite|improve this answer
Thanks! Noticed my mistake. – Aspiring Mathlete May 2 '14 at 12:10

You could start with $a+b$ and $ab$,

  1. Note first $(a+b)^2 - 2ab = a^2+ b^2 $

  2. Note that $(a^2+b^2)^2 - 2(ab)^2 = a^4 + b^4$

As an asside, i use this set of numbers to factorise $b^n-a^n$. Part of the process involves creating a sequence $T_n = a^n+b^n$. You find that $T_{n+1}=(a+b)T_n - ab T_{n-1}$

Applying $T_0=2$ and $T_1$=5 we get this series for the sum of $a^n+b^n$ for n=0 t0 6. The iteration is t(n+1)=5t(n)-2t(n-1).

      0    1    2    3    4     5      6
      2    5   21   95  433  2070   9484

Doing it this way, allows one to evaluate symmetric equations (ie $f(a,b)=f(b,a)), very quickly. You start at the middle, and work outwards, multiplying the result by ab, before adding the next term.

share|cite|improve this answer

There is a general recursive identity for such questions: let $f(n) = a^n + b^n$. Then $$f(n+1) = f(1)f(n) - ab f(n-1).$$ Note that $f(0) = 2$ provided $ab \ne 0$. Then in your particular case, we wish to find $f(4)$, where $$f(n+1) = 5f(n) - 2f(n-1).$$ With starting values $f(0) = 2$ and $f(1) = 5$, we easily compute $f(2) = 5(5)-2(2) = 21$, $f(3) = 5(21) - 2(5) = 95$, $f(4) = 5(95) - 2(21) = 433$. The advantage of this approach is that it can be used to compute sums of higher powers quite easily. It also leads to a general solution via the solution of the associated linear recurrence; e.g., with generating functions.

share|cite|improve this answer

An easier approach: Since $ab = 2$, then

$$ (a+b)^2 = 25 $$. Hence

$$ a^2 + b^2 + 2ab = 25 \iff a^2 + b^2 = 25 - 2 ab = 21 $$


$$ (21)^2 = (a^2 + b^2)^2 = a^4 + b^4 + 2 a^2 b^2 = a^4 + b^4 + 2(2)^2 $$

share|cite|improve this answer

Another totally different approach is tho solve for $a$ and $b$ using : $$ab = 2, \qquad a+b = 5$$ This is like solving a second order polynomial $(x+a)(x+b) = x^2 + (a+b)x+ab$ where the roots are $-a$ and $-b$.

So you could solve the polynomial $x^2 + 5x + 2 = 0$, deduce $a$ and $b$ and calculate $a^4+b^4$.

share|cite|improve this answer
I might be totally wrong, but it seems to me that the whole point of the exercise is precisely to avoid computing the exact values of $a, b$, which involve a square root, and use only their sum and product, which are integers. – Andreas Caranti May 2 '14 at 11:04
@AndreasCaranti: Possibly indeed but the OP didn't specify so I thought I'll give another option. – user88595 May 2 '14 at 11:06

Use $a+b=5$ to write $b$ in terms of $a$: $$b=5-a$$ Then substitute this for b in the other equation and solve for $a$: $$a(\overbrace{5-a}^{b}) =2\iff 5a-a^2=2\iff a^2-5a+2=0\iff a=\frac52\pm\frac12\sqrt{17}$$ (using the quadratic formula). Since the original equations were symmetric in $a$ and $b$ (you could have solved for $b$ instead and arrived at exactly the same values), these two numbers (the $\pm$) are the values of $a$ and $b$: $$a=\frac52\pm\frac12\sqrt{17} \textrm{ and } b=\frac52\mp\frac12\sqrt{17}$$

Since this is homework, I will leave it up to you to compute that $$a^4+b^4 = \boxed{433}$$ given these values of $ a$ and $b$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.