# Can this equation have four integer solutions?

I am trying to find three numbers a, b, c so that the equation $$(x+a)^4 + (x+b)^4 = c$$ have four integer solutions. I tried

f[x_] := (x + a)^4 + (x + b)^4 - c
Eliminate[{f[x] == 0 // PowerExpand, x + (a + b)/2 == t}, x] // FullSimplify


(a - b)^4 + 24 (a - b)^2 t^2 + 16 t^4 == 8 c

And then, I solve

Solve[(a - b)^4 + 24 (a - b)^2 t^2 + 16 t^4 - 8 c == 0, t] // FullSimplify


{{t -> -(1/2) Sqrt[-3 (a - b)^2 - 2 Sqrt[2] Sqrt[(a - b)^4 + c]]}, {t -> 1/2 Sqrt[-3 (a - b)^2 - 2 Sqrt[2] Sqrt[(a - b)^4 + c]]}, {t -> -(1/ 2) Sqrt[-3 (a - b)^2 + 2 Sqrt[2] Sqrt[(a - b)^4 + c]]}, {t -> 1/2 Sqrt[-3 (a - b)^2 + 2 Sqrt[2] Sqrt[(a - b)^4 + c]]}}

Now I put

r1 = -(1/2) Sqrt[-3 (a - b)^2 - 2 Sqrt[2] Sqrt[(a - b)^4 + c]] - (a + b)/2;
r2 = 1/2 Sqrt[-3 (a - b)^2 - 2 Sqrt[2] Sqrt[(a - b)^4 + c]] - (a + b)/
2;
r3 = -(1/2) Sqrt[-3 (a - b)^2 +
2 Sqrt[2] Sqrt[(a - b)^4 + c]] - (a + b)/2;
r4 = 1/2 Sqrt[-3 (a - b)^2 + 2 Sqrt[2] Sqrt[(a - b)^4 + c]] - (a + b)/
2;

• This sequence is of relevance oeis.org/A003824 Commented Jun 11, 2023 at 14:47
• Do $a$, $b$, and $c$ have to be real? Positive? Integers? All of the above? Commented Jun 12, 2023 at 20:02
• Also note that the first two of your solutions for $t$ are pretty obviously going to be complex numbers if $a$, $b$, and $c$ are real. If that's the case there's no way that you're going to get four distinct integer solutions. Commented Jun 12, 2023 at 20:06
• Another way to see this is that if you define $f(x) = (x+a)^4 + (x+b)^4 - c$, then $f''(x) = 12(x+a)^2 + 12(x+b)^2 \geq 0$. This means that the graph is concave up everywhere, and so $f(x)$ cannot have more than two real roots. Commented Jun 12, 2023 at 20:20

Your equation has a maximum of two real solutions, so having four integer solutions is out of the question:

Solve[f[x]==0, x, Reals]

{{x -> ConditionalExpression[
Root[ (* original polynomial *), 1],
a^4 - 4 a^3 b + 6 a^2 b^2 - 4 a b^3 + b^4 - 8 c < 0]},
{x -> ConditionalExpression[
Root[ (* original polynomial *), 2],
a^4 - 4 a^3 b + 6 a^2 b^2 - 4 a b^3 + b^4 - 8 c < 0]}}


In other words, there are two real roots if $$(a-b)^4 - 8 c < 0$$, and no real roots otherwise.

To see why this is using high-school algebra techniques, apply the transformation $$x = \sqrt{u}-\frac12(a+b)$$:

FullSimplify[Expand[f[x] /. {x -> Sqrt[u] - (a + b)/2}]]

(* 1/8 (a - b)^4 - c + 3 (a - b)^2 u + 2 u^2 *)


Each of the two roots $$u_\pm$$ of this polynomial will correspond to two roots of the original polynomial. If we want real roots for $$x$$, then both the roots for $$u$$ must be real and positive. But the sum of the roots of this latter polynomial is $$-3(a-b)^2/2 < 0$$. So at least one solution for $$u$$ must be negative, meaning that there is at most one positive root for $$u$$ and therefore at most two real roots for $$x$$.

We can ensure that the other $$u$$ root is positive by noting that the product of the roots of the quadratic polynomial is $$(\frac18(a - b)^4 - c)/2$$. Since one root is negative, and we want the other root to be positive, this quantity must be negative.

Try

solx = Values@Solve[(x + a)^4 + (x + b)^4 - c == 0, x] // Flatten
cond=Reduce[Element[solx,Integers],{a,b,c},Reals]


Forcing the last two conditions we find

cond /. c -> -a^4 + 4 a^3 b - 6 a^2 b^2 + 4 a b^3 - b^4 /. b -> a
(*(a | a | a | a) \[Element] Integers*)


solution a=b=integer,c=0

• I would interpret the original question as seeking 4 different integer roots, not a single root repeated 4 times (which I think you have). Commented Jun 11, 2023 at 17:05
• Agrred, but result of my evaluation gives 4 identical roots. Commented Jun 11, 2023 at 17:53

We can solve $$(x+a)^4 + (x+b)^4 = c$$ for integer values. The result is a ConditionalExpression with one integer constant.

Clear[a,b,c,x]
sol = First[Solve[(x+a)^4 + (x+b)^4 == c, x, Integers]];

{x -> ConditionalExpression[C[1],
Element[a, Integers] && Element[b, Integers] &&
Element[C[1], Integers] &&
c == a^4 + b^4 + 4*a^3*C[1] + 4*b^3*C[1] + 6*a^2*C[1]^2 +
6*b^2*C[1]^2 + 4*a*C[1]^3 + 4*b*C[1]^3 + 2*C[1]^4]}


Let's use sol to find values for x, a, b, and c.

TableForm[With[{xx = 1}, (* x, a, and b are any integers *)
Flatten[Table[{xx, i, j, (x/.sol/.{a->i, b->j, C[1]->xx})[[2,2]]},
{i, -2, 2}, {j, -2, 2}], 1]],

x a b c
1 -2 -2 2
1 -2 -1 1
1 -2 0 2
...
1 2 0 82
1 2 1 97
1 2 2 162

Another way to see the results is to compute c from values of a and b and plot the result.

ListPlot3D[Flatten[
Table[{i, j, (x/.sol/.{a->i, b->j, C[1]->1})[[2,2]]},
{i, -10, 10}, {j, -10, 10}], 1],
PlotRange -> All, AxesLabel -> {a,b}, BaseStyle -> {Bold, 14}]


We can make a list of random solutions.

SeedRandom[1357];
numSamples = 5;
TableForm[
Sort@Flatten[Last[Reap[Do[
Sow[{xx, a, b, Last[(x/.sol)/.C[1]->xx/.Equal->List][[-1, -1]]}/.
{a->#1, b->#2, xx->#3}&@@RandomInteger[{-9, 9}, 3]],numSamples]]], 1],

x a b c
-9 9 -3 20736
-7 -4 -1 18737
-2 -1 5 162
2 3 2 881
9 -5 0 6817

Check that x, a, b, and c are a valid solution.

checkTrueQ[{x_, a_, b_, c_}] := (x + a)^4 + (x + b)^4==c
checkTrueQ[{-9, 9, -3, 20736}]
(*True*)

• If I'm reading your code correctly you've found a way to generate values of $a$, $b$, and $c$ so that the equations $(x+a)^4 + (x+b)^4 = c$ has one integer solution. But the OP wanted this equation to have four (presumably distinct) integer solutions. Commented Jun 12, 2023 at 19:55
• @MichaelSeifert Yes, that's what I've done. That's how I understand John's question. I see that your answer says there can't be four simultaneous solutions. Maybe John will explain his goal. Commented Jun 13, 2023 at 0:15